Question 15 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4n-1)(4n+3)}}=\frac{\text{n}}{3(\text{4n}+3)}$
AnswerLet P(n): $\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4n-1)(4n+3)}}=\frac{\text{n}}{3(\text{4n}+3)}$
For n = 1
$\frac{1}{3.7}=\frac{3}{(7)}$
$\frac{1}{21}=\frac{1}{21}$
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4k-1)(4k+3)}}=\frac{\text{k}}{3(\text{4k}+3)} \ ...(1)$
We have to show that
$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4k-1)(4k+3)}}+\frac{1}{(4\text{k}+3)(4\text{k}+7)}=\frac{(\text{k}+1)}{3\text{(4k}+7)}$
Now,
$\Big\{\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4k-1)(4k+3)}}\Big\}+\frac{1}{(4\text{k}+3)(4\text{k}+7)}$
$=\frac{\text{k}}{3(4\text{k}+3)}+\frac{1}{(4\text{k}+3)(4\text{k}+7)}$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{\text{k}}{3}+\frac{1}{(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{\text{k}(4\text{k}+7)+3}{3(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{4\text{k}^2+7\text{k}+3}{3(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{4\text{k}^2+4\text{k}+3\text{k}+3)}{3(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{(4\text{k}(\text{k}+1)+3(\text{k}+1)}{3(4\text{k}+7)}\Big]$
$=\frac{1}{(4\text{k}+3)}+\Big[\frac{(4\text{k}+3)(\text{k}+1)}{3(4\text{k}+7)}\Big]$
$=\frac{(\text{k}+1)}{3(4\text{k}+7)}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N} $ by PMI
View full question & answer→Question 25 Marks
Prove the following by the principle of mathematical induction:
$1 + 3 + 3^2 + ... + 3^{\text{n}-1}=\frac{3^\text{n}-1}{2}$
AnswerLet P(n) be the given statement. Now, $\text{P(n)}=1 + 3 + 3^2 + ... + 3^{\text{n}-1}=\frac{3^\text{n}-1}{2}$
Step 1:
$\text{P(1)}=1 =\frac{3^1-1}{2}=\frac{2}{2}=1$ Hence, P(1) is true.
Step 2:
Let P(m) is true. Then, $1 + 3 + 3^2 + ... + 3^{\text{m}-1}=\frac{3^\text{m}-1}{2}$ We shall prove that P(m + 1) is true. That is, $1 + 3 + 3^2 + ... + 3^{\text{m}}=\frac{3^\text{m+1}-1}{2}$ Now, we have: $1 + 3 + 3^2 + ... + 3^{\text{m-1}}=\frac{3^\text{m}-1}{2}$ $\Rightarrow1 + 3 + 3^2 + ... + 3^{\text{m-1}}+3\text{m}=\frac{3^\text{m}-1}{2}+3\text{m}$ [Adding $3^m$ to both sides] $\Rightarrow1 + 3 + 3^2 + ... +3\text{m}=\frac{3^\text{m}-1+2\times3^\text{m}}{2}=\frac{3^\text{m}(1+2)-1}{2}=\frac{3^\text{m+1}-1}{2}$ Hence, P(m + 1) is true. By the principle of mathematical induction, P(n) is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 35 Marks
Prove that $\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+...+\cos(\alpha+(\text{n}-1)\beta)\\=\frac{\cos\Big\{\alpha+\big(\frac{\text{n}-1}{2}\big)\beta\Big\}\sin\big(\frac{\text{n}\beta}{2}\big)}{\sin\frac{\beta}{2}}$ For all $\text{n}\in\text{N}.$
AnswerLet
P(n): $\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+...+\cos(\alpha+(\text{n}-1)\beta]\\=\frac{\cos\Big\{\alpha+\big(\frac{\text{n}-1}{2}\big)\beta\Big\}\sin\big(\frac{\text{n}\beta}{2}\big)}{\sin\frac{\beta}{2}}$ For all $\text{n}\in\text{N}.$
Step 1: For n = 1
$\text{LHS }=\cos[\alpha(1-1)\beta]=\cos\alpha$
$\text{RHS }=\frac{\cos\Big\{\text{a}+\big(\frac{1-1}{2}\big)\beta\Big\}\sin\big(\frac{\beta}{2}\big)}{\sin\big(\frac{\beta}{2}\big)}=\cos\alpha$
As, LHS = RHS
So, it is true for n = 1.
Step II: For n = k
Let p(k): $\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ ...\ +\cos[\alpha+(\text{k}-1)\beta]=\frac{\cos\big\{\alpha+\big(\frac{\text{k}-1}{2}\big)\beta\big\}\sin\big(\frac{\text{k}\beta}{2}\big)}{\sin\big(\frac{\beta}{2}\big)}$ be true
Step III: For n = k + 1,
$\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+....+\cos(\alpha+(\text{k}-1)\beta]+\cos[\alpha+(\text{k}+1-1\beta)$
$=\frac{\cos\bigg\{\alpha+\Big(\frac{\text{k}-1}{2}\Big)\beta\sin\Big(\frac{\text{k}\beta}{2}\Big)}{\sin\Big(\frac{\beta}{2}\Big)}+\cos(\text{a}+\text{k}\beta)$
$=\frac{\cos\bigg\{\alpha+\Big(\frac{\text{k}-1}{2}\Big)\beta\bigg\}\sin\Big(\frac{\text{k}\beta}{2}\Big)+\sin\big(\frac{\beta}{2}\big)\cos(\text{a}+\text{k}\beta)}{\sin\frac{\beta}{2}}$
$=\frac{\sin\big(\alpha+\text{k}\beta-\frac{\beta}{2}\big)-\sin\big(\alpha-\frac{\beta}{2}\big)+\sin\big(\text{a}+\text{k}\beta+\frac{\beta}{2}\big)-\sin\big(\text{a}+\text{k}\beta-\frac{\beta}{2}\big)}{2\sin\big(\frac{\beta}{2}\big)}$
$=\frac{-\sin\big(\alpha-\frac{\beta}{2}\big)\sin\big(\frac{\text{k}\beta+\beta}{2}\big)}{2\sin\big(\frac{\beta}{2}\big)}$
$=\frac{2\cos\big(\frac{2\text{a}+\text{k}\beta}{2}\big)\sin\big(\frac{\text{k}\beta+\beta}{2}\big)}{2\sin\big(\frac{\beta}{2}\big)}$
$=\frac{2\cos\big(\frac{2\text{a}+\text{k}\beta}{2}\big)\sin\big(\frac{(\text{k}+1)+\beta}{2}\big)}{2\sin\big(\frac{\beta}{2}\big)}$
$\text{RHS}=\frac{\cos\Big\{\alpha+\big(\frac{\text{k}+1-1}{2}\big)\beta\Big\}\sin\big(\frac{(\text{k}+1\beta)}{2}\big)}{\sin\big(\frac{\beta}{2}\big)}$
$=\frac{\cos\big(\alpha+\frac{\text{k}\beta}{2}\big)\sin\big(\frac{(\text{k}+1\beta)}{2}\big)}{\sin\big(\frac{\beta}{2}\big)}$
As, LHS = RHS
So, it is also true for n = k + 1.
View full question & answer→Question 45 Marks
Prove the following by the principle of mathematical induction:
$1.3 + 3.5 + 5.7 + ... + (2\text{n} - 1)(2\text{n} + 1)= \frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$
AnswerLet P(n): $1.3 + 3.5 + 5.7 + ... +(2\text{n} - 1)(2\text{n} + 1)= \frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$
for n = 1
$1.3=\frac{1(4+6-1)}{3}$
3 = 3
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$1.3 + 3.5 + 5.7 + ... +(2\text{k} - 1)(2\text{k} + 1)= \frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3} \ ...(1)$
We have to show that
$1.3 + 3.5 + 5.7 + ... +(2\text{k} - 1)(2\text{k} + 1)+(2\text{k} + 1)(2\text{k} + 3) $
$=\frac{(\text{k}+1)\big[4(\text{k} + 1)^2+6(\text{k+1})-1\big]}{3}$
Now,
${1.3 + 3.5 + 5.7 + ... + (2\text{k} - 2)(2\text{k} + 1)} + (2\text{k} + 1)(2\text{k} + 3)$
$=\frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3}+(2\text{k}+1)(2\text{k}+3)$ [Using equation (1)]
$=\frac{\text{k}(4\text{k}^2+6\text{k}-1)+3(4\text{k}^2+6\text{k}+2\text{k}+3)}{3}$
$=\frac{4\text{k}^3+6\text{k}-\text{k}+12\text{k}^2+18\text{k}+6\text{k}+9}{3}$
$=\frac{4\text{k}^3+18\text{k}^2+23\text{k}+9}{3}$
$=\frac{4\text{k}^3+4\text{k}^2+14\text{k}^2+14\text{k}+9\text{k}+9}{3}$
$=\frac{(\text{k}+1)(4\text{k}^2+8\text{k}+4+6\text{k}+6-1)}{3}$
$=\frac{(\text{k}+1)\big[(4(\text{k}+1)^2+6(\text{k}+1)-1\big]}{3}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 55 Marks
Prove that $\frac{(2\text{n})!}{2^{2\text{n}}(\text{n}!)^2}\leq\frac{1}{\sqrt{3\text{n}+1}}$ for all $\text{n}\in\text{N}.$
AnswerP(n): $\frac{(2\text{n})!}{2^{2\text{n}}(\text{n}!)^2}\leq\frac{1}{\sqrt{3\text{n}+1}}$
For n = 1
$\frac{2!}{2^2.1}\leq\frac{1}{\sqrt{4}}$
$=\frac{1}{2}\leq\frac{1}{2}$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, So
$\frac{(2\text{k})!}{2^{2\text{k}}(\text{k}!)^2}\leq\frac{1}{\sqrt{3\text{k}+1}} \ ...(1)$
We have to show that,
$\frac{2(\text{k+1})!}{2^{2(\text{k+1})}\big[(\text{k+1})!\big]^2}\leq\frac{1}{\sqrt{3\text{k}+4}}$
Now,
$\frac{2(\text{k+1})!}{2^{2(\text{k+1})}\big[(\text{k+1})!\big]^2}$
$=\frac{(2\text{k+2})!}{2^{2\text{k}}.2^{2}(\text{k+1})!{(\text{k+1})!}}$
$=\frac{(2\text{k+2})(2\text{k+1})(2\text{k})!}{4.2^{2}(\text{k+1})(\text{k}!){(\text{k+1})(\text{k}!)}}$
$=\frac{2(\text{k+2})(2\text{k+1})(2\text{k})!}{4.(\text{k+1})^2.2^{2\text{k}}.(\text{k}!)}^2$
$\leq\frac{2(2\text{k+1})}{4(\text{k}+1)}.\frac{1}{\sqrt{3\text{k}+1}}$
$\leq\frac{(2\text{k+1})}{2(\text{k}+1)}.\frac{1}{\sqrt{3\text{k}+1}}$
$\leq\frac{(2\text{k+1})}{2(\text{k}+1)}.\frac{1}{\sqrt{3\text{k}+3+1}}$
$\leq\frac{1}{\sqrt{3\text{k}+4}} \ \begin{bmatrix}\text{Since,} \ 2 \text{k}+2<2\text{k}+2\\\ 3\text{k}+1\leq 3\text{k}+4\end{bmatrix}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 65 Marks
Prove the following by the principle of mathematical induction:
1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! - 1 for all $\text{n}\in\text{N}$
AnswerConsider equation
1 × 1! + 2 × 2! + 3 × 3! + ... + n × n!
Lets take (n + 1)! -n! = n!(n + 1 - 1) = n × n!
Now substitue n = 1, 2, 3, 4 ... n in above equation we get
2! - 1! = 1 × 1!
3! - 2! = 2 × 2!
4! - 3! = 3 × 3!
.....
(n + 1)! -n! = n × n!
Adding all the above terms gives
1 × 1! + 2 × 2! + 3 × 3! + ... n × n! = 2! - 1! + 3! - 2! + 4! - 3! ... (n + 1)! - n!
1 × 1! + 2 × 2! + 3 × 3! + ... n × n! = (n + 1)! - 1
View full question & answer→Question 75 Marks
A sequence $a_1, a_2, a_3, \ldots$ is defined by letting $a_1=3$ and $a_k=7 a_k-1$ for all natural numbers $k \geq 2$. Show that $a_n=$ $3.7^{ n -1}$ for all $n \in N$.
AnswerLet p(n) be the statement given by
$P(n): a_n = 3 \times 7^{n-1} $ for all $\text{n}\in\text{N}.$
Step $1:$
$p(2): a_2 = 3 \times 7^{2-1} = 21$
Given that $a_k = 7a_{k-1}$ for all natural numbers $\text{k}\geq2$
$a_2 = 7a_1 = 7 \times 3 = 21$
$\therefore$ p(2) is true.
Step 2:
Let p(m) is true. Then,
$a_m = 3 \times _7^{m-1} ... (1)$
We have to prove that p(m + 1) is true.
$a_{m+1} = 7a_m$
$a_{m+1} = 7 \times a_m$
$a_{m+1} = 7^1 \times 3 \times 7^{m-1} ...$ [From $(1)$]
$a_{m+1} = 3 \times 7^{m-1+1}$
$a_{m+1} = 3 \times 7^m$
$\Rightarrow p(m + 1)$ is true.
Hence by the principle of mathematical induction, the given results is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 85 Marks
Prove the following by the principle of mathematical induction:
$5^{2n+2} + 24n - 25$ is divisible by $576$ for all $\text{n}\in\text{N}$
AnswerLet $p(n): 5^{2n+2} + 24n - 25 $is divisible by $576$ for all $576$
For $n = 1$
$5^4 - 24 - 25$
$= 625 - 49$
$= 576$
Which is divisible by 576
$\Rightarrow p(n)$ is true for $n = 1$
Let p(n) is true for $n = k$, so
$5^{2k+2} - 24k - 25$ is divisible by 576
$5^{2\text{k}+2} - 24\text{k} - 25=576\lambda \ ...(1)$
We have to show that,
5^(^{2k+2)+2} - 24(k+1) - 25 is divisible by 576
$5^{(2\text{k}+2)+2} - 24(\text{k}+1) - 25=576\mu$
Now,
$5^{(2\text{k}+2)+2} - 24(\text{k}+1) - 25$
$=5^{(2\text{k}+2)}.5^2 - 24\text{k}-24-25$
$(576\lambda+24\text{k}+25)25-24\text{k}-49$ [Using equation (1)]
$=25.576\lambda+600\text{k}+625-24\text{k}-49$
$=25.576\lambda+576\text{k}+576$
$=576(25\lambda+\text{k}+1)$
$=576\mu$
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 95 Marks
Prove the following by the principle of mathematical induction:
$1 + 3 + 5 + ... + (2n - 1) = n^2 $i.e., the sum of first n odd natural numbers is $n^2.$
AnswerLet $P(n): 1+3+5+\ldots+(2 n-1)=n^2$
For $n=1$
$P(1): 1=1^2$
$1=1$
$\Rightarrow P(n)$ is true for $n=1$
Let $P(n)$ is true for $n=k$, so
$P(k): 1+3+5+\ldots+(2 k-1)=k^2 \ldots$
We have to show that
$1+3+5+\ldots+(2 k-1)+2(k+1)-1=(k+1)^2$
Now,
${1 + 3 + 5 + ... + (2k - 1)} + 2(k + 1)$
$= k^2 + 2(k + 1)$ [Using equation $(1)]$
$= k^2 + 2k + 1$
$= (k + 1)^2$
$\Rightarrow P ( n )$ is true for $n = k +1$
$\Rightarrow P(n)$ is true for all $n \in N$ by $PMI$
View full question & answer→Question 105 Marks
Prove that $\frac{1}{\text{n}+1}+\frac{1}{\text{n}+2}+...+\frac{1}{2\text{n}}>\frac{13}{24},$ For all natural numbers n > 1.
Answer$\frac{1}{\text{n}+1}+\frac{1}{\text{n}+2}+...+\frac{1}{2\text{n}}>\frac{13}{24},$
Using induction we first show this is true for n = 2.
$\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}>\frac{13}{24}$ (True)
Now lets assume it is true for some n = k,
$\text{S}_\text{k}=\frac{1}{\text{k}+1}+\frac{1}{\text{k}+2}+...+\frac{1}{2\text{k}}>\frac{13}{24}$
Finally we need to prove that this implies it is also true for n = k + 1:
$\text{S}_\text{k+1}=\frac{1}{\text{k}+2}+\frac{1}{\text{k}+3}+...+\frac{1}{2\text{k}+2}$
$=\frac{-1}{\text{k}+1}+\frac{1}{\text{k}+1}+\frac{1}{\text{k}+2}+\frac{1}{\text{k}+3}+...+\frac{1}{2\text{k}}+\frac{1}{2\text{k}+1}+\frac{1}{2\text{k}+2}$
$=\frac{-1}{\text{k}+1}+\text{S}_\text{k}+\frac{1}{2\text{k}+1}+\frac{1}{2\text{k}+2}$
$=\text{S}_\text{k}+\frac{1}{2(2\text{k}+1)(\text{k}+1)}$
$>\text{S}_\text{k}$
$\therefore\text{S}_\text{k+1}>\frac{13}{24}$
View full question & answer→Question 115 Marks
Prove the following by the principle of mathematical induction:
$3^{2n} + 7$ is divisible by $8$ for all $\text{n}\in\text{N}$
AnswerLet $p(n): 3^{2n} + 7$ is divisible by $8$ for all $\text{n}\in\text{N}$
For $n = 1$
$3^2 + 7 = 16$
Which is divisible by $8$
$\Rightarrow p(n)$ is true for $n = 1$
Let p(n) is true for $n = k$, so
$3^{2k} + 7$ is divisible by $8$
$\Rightarrow3^{2\text{k}} + 7 =8\lambda \ ...(1)$
We have to show that,
$3^{2(k+1)} + 7$ is divisible by $8$
$3^{(2\text{k+1})} + 7 =8\mu$
Now,
$3^{2(k+1)} + 7$
$= 3^{2k}.3^2 + 7$
$= 9.3^{2k} + 7$
$=9.(8\lambda-7)+7$
$=72\lambda-56$
$=8(9\lambda-7)$
$=8\mu$
$\Rightarrow p(n)$ is true for $n = k + 1$
$\Rightarrow p(n)$ is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 125 Marks
Prove the following by the principle of mathematical induction:
$1.3 + 2.4 + 3.5 + ... + \text{n}(\text{n} + 2) $
$=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+7)$
AnswerLet $1.3 + 2.4 + 3.5 + ... + \text{n}(\text{n} + 2) $
$=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+7)$
for n = 1
$1.3=\frac{1}{6}.1.(2)(9)$
3 = 3
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$1.3 + 2.4 + 3.5 + ... + \text{k}(\text{k} + 2) =\frac{1}{6}\text{k}(\text{k}+1)(2\text{k}+7) \ ...(1)$
We have to show that
$1.3 + 2.4 + 3.5 + ... + \text{k}(\text{k} + 2)+(\text{k} + 1)(\text{k} + 3)$
$ =\frac{\text{k}+1}{6}(\text{k}+2)(2\text{k}+9)$
Now,
{1.3 + 2.4 + 3.5 + ... + k(k + 2)} + (k + 1)(k + 3)
$ =\frac{1}{6}\text{k}(\text{k}+1)(2\text{k}+7)+(\text{k}+1)(\text{k}+3)$ [Using equation (1)]
$(\text{k} + 1)\Big[\frac{\text{k}(2\text{k}+7)}{6}+\frac{\text{k}+3}{1}\Big]$
$(\text{k} + 1)\Big[\frac{2\text{k}^2+7\text{k}+6\text{k}+18}{6}\Big]$
$(\text{k} + 1)\Big[\frac{2\text{k}^2+13\text{k}+18}{6}\Big]$
$(\text{k} + 1)\Big[\frac{2\text{k}^2+4\text{k}+9\text{k}+18}{6}\Big]$
$(\text{k} + 1)\Big[\frac{2\text{k}+(\text{k}+2)+9(\text{k}+2)}{6}\Big]$
$(\text{k} + 1)\Big[\frac{(2\text{k}+9)(\text{k}+2)}{6}\Big]$
$\frac{1}{6}(\text{k}+1)(\text{k}+2)(2\text{k}+9)$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 135 Marks
Prove that $\Big(1-\frac{1}{2^2}\Big)\Big(1-\frac{1}{3^2}\Big)\Big(1-\frac{1}{4^2}\Big)...\Big(1-\frac{1}{\text{n}^2}\Big)=\frac{\text{n}+1}{2\text{n}}$ for all natural numbers, $\text{n}\geq2.$
Answer$\Big(1-\frac{1}{2^2}\Big)\Big(1-\frac{1}{3^2}\Big)\Big(1-\frac{1}{4^2}\Big)...\Big(1-\frac{1}{\text{n}^2}\Big)$
Above can be written as
$=\Big(\frac{2^2-1}{2^2}\Big)\Big(\frac{3^2-1}{3^2}\Big)\Big(\frac{4^2-1}{4^2}\Big)...\Big(\frac{\text{n}^2-1}{\text{n}^2}\Big)$
$=\Big(\frac{(2+1)(2-1)}{2^2}\Big)\Big(\frac{(3+1)(3-1)}{3^2}\Big)\Big(\frac{(4+ 1)(4-1)}{4^2}\Big)...\Big(\frac{(\text{n}+1)(\text{n}-1)}{\text{n}^2}\Big)$
$=\Big(\frac{3.1}{2^2}\Big)\Big(\frac{4.2}{3^2}\Big)\Big(\frac{5.3}{4^2}\Big)...\Big(\frac{(\text{n}+1)(\text{n}-1)}{\text{n}^2}\Big)$
In the above product, there are two series in numerator 3.4.5...(n + 1) and 1.2.3...(n - 1). All numbers from 3 to (n-1) are repeated twice and 1,2,n are appeared once in numerator. So after cancelling like terms we get
$=\frac{\text{n}+1}{2\text{n}}$
View full question & answer→Question 145 Marks
Prove the following by the principle of mathematical induction:
$1 + 2 + 3 + ... + \text{n}=\frac{\text{n}(\text{n}+1)}{2}$ i.e, the sum of the first n natural numbers is $\frac{\text{n}(\text{n}+1)}{2}.$
AnswerLet P(n): $1 + 2 + 3 + ... + \text{n}=\frac{\text{n}(\text{n}+1)}{2}$
For n = 1,
LHS of P(n) = 1
RHS of P(n) $=\frac{\text{1}(\text{1}+1)}{2}1=1$
Since, LHS = RHS
⇒ P(n) is true for n = 1
Let P(n) be true for n = k, so
$1 + 2 + 3 + ... + \text{k}=\frac{\text{k}(\text{k}+1)}{2}...(1)$
Now,
$ (1 + 2 + 3 + ... + \text{k}) + (\text{k} + 1)$
$=\frac{\text{k}(\text{k}+1)}{2}+(\text{k}+1)$
$=(\text{k}+1)\Big(\frac{\text{k}}{2}+1\Big)$
$=\frac{(\text{k}+1)(\text{k}+2)}{2}$
$=\frac{(\text{k}+1)[(\text{k}+1)+1]}{2}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$
So, by the principle of mathematical induction
P(n): $ 1 + 2 + 3 + ... + \text{n}=\frac{\text{n}(\text{n}+1)}{2}$ is ture for all $\text{n}\in\text{N}$
View full question & answer→Question 155 Marks
Prove the following by the principle of mathematical induction:
$11^{n+2} + 12^2^{n+1}$ is divisible of $133$ for all $\text{n}\in\text{N}$
AnswerLet p(n): $11^{n+2} + 12^2^{n+1}$ is divisible of 133
For $n = 1$
$11^3 + 12^3$
$= 1331 + 1728$
$= 3059$
It is divisible of $133$
$\Rightarrow p(n)$ is true for $n = 1$
Let p(n) is true for $n = k$, so
$11^{k+2} + 12^2^{k+1}$ is divisible of 133
$11^{\text{k}+2}+12^{2\text{k}+1}=133\lambda \ ...(1)$
We have to show that,
$1^{k+3} + 12^2^{k+3} $ is divisible of $133$
Now,
$11^{\text{k}+2}.11+12^{2\text{k}+1}.12^2$
$= (133\lambda-12^{2\text{k}+1})11+12^{2\text{k}+1}.144$
$=11.133\lambda+11.12^{2\text{k}+1}+144.12^{2\text{k}+1}$
$=11.133\lambda+133.12^{2\text{k}+1}$
$=133(11\lambda+12^{2\text{k}+1})$
$=133\mu$
$\Rightarrow p(n)$ is true for $n = k + 1$
$\Rightarrow p(n)$ is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 165 Marks
Prove the following by the principle of mathematical induction:
$1^2 + 2^2 + 3^2 +...+ \text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
AnswerLet P(n): $1^2 + 2^2 + 3^2 + ... +\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
For $n = 1$
$\text{P}(1):1=\frac{1(1+1)(2+1)}{6}$
$1=1$
$\Rightarrow P(n)$ is true for $n = 1$
Let $P(n)$ is true for $n = k$, so
P(k): $1^2 + 2^2 + 3^2 + .... + \text{k}^2=\frac{\text{k}(\text{k}+1)(2\text{k}+1)}{6}\ \cdots(1)$
We have to show that P(n) is true for $n = k + 1$
$\Rightarrow 1^2 + 2^2 + 3^2 + .... + k^2 + (k + 1)^2 $$=\frac{(\text{k}+1)(\text{k}+2)(2\text{k}+3)}{6}$
So, $1^2 + 2^2 + 3^2 + .... + k^2 + (k + 1)^2$^
$=\frac{\text{k}(\text{k}+1)(2\text{k}+1)}{6}+(\text{k}+1)^2$ [Using equation (1)]
$=(\text{k}+1)\Big[\frac{2\text{k}^2+\text{k}}{6}+\frac{(\text{k}+1)}{1}\Big]$
$=(\text{k}+1)\Big[\frac{2\text{k}^2+\text{k}+6\text{k}+6}{6}\Big]$
$=(\text{k}+1)\Big[\frac{2\text{k}^2+7\text{k}+6}{6}\Big]$
$=(\text{k}+1)\Big[\frac{2\text{k}^2+4\text{k}+3\text{k}+6}{6}\Big]$
$=(\text{k}+1)\Big[\frac{2\text{k(k+2)+3(k+2)}}{6}\Big]$
$=\frac{(\text{k}+1)(2\text{k+3})(\text{k}+2)}{6}$
$\Rightarrow P(n)$ is true for $n = k + 1$
$\Rightarrow P(n) $is true for all $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 175 Marks
Prove that the number of subsets of a set containing n distinct elements is $2^n$^ for all $\text{n}\in\text{N}.$
AnswerLet $p(n)$ be the statement given by
$P(n)$ : The number of subsets of a set containing $n$ distinct element is is $2^n$ for all $n \in N$.
Step 1:
$P(1): 2^1=2$
For any set A containing 1 element, empty set and set A are two sets always subsets of A . $\therefore P (1)$ is true.
Step 2:
Let $p(m)$ is true. Then,
A set containing $m$ distinct elements has $2^m$ subsets ...
We have to prove that $p(m+1)$ is true.
Let the set $A$ has $(m+1)$ elements.
$A=\{1,2 \ldots, m, m+1\}$
$A=\{1,2 \ldots, m\} \cup\{m+1\}$
Now using (1) we can say that $\{1,2, \ldots, m\}$ being $m$ elements has $2^m$ subsets
For $\{m+1\}$, empty set and itself $\{m+1\}$ are subsets
So, $\{m+1\}$ has 2 subsets
$\Rightarrow$ Set $A$ has $2^m+2$ subsets
$\Rightarrow$ Set $A$ has $2^{m+1}$ subsets
$\Rightarrow p ( m +1)$ is true.
Hence by the principle of mathematical induction, the given results is true for all $n \in N$.
View full question & answer→Question 185 Marks
Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.
AnswerLet P(n) be the statement 3n < n!
For n = 1,
3n = 3 × 1 = 3
n! = 1! = 1
Now, 3 > 1
So, P(1) is not true.
For n = 2,
3n = 3 × 2 = 6
n! = 2! = 2
Now, 6 > 2
So, P(2) is not true.
For n = 3,
3n = 3 × 3 = 9
n! = 3! = 6
Now, 9 > 6
So, P(3) is not true.
For n = 4,
3n = 3 × 4 = 12
n! = 4! = 24
Now, 12 < 24
So, P(4) is true.
For n = 5,
3n = 3 × 5 = 15
n! = 5! = 120
Now, 15 < 120
So, P(5) is true.
Similarly, it can be verified that 3n < n! for n = 6, 7, 8, ...
Thus, the statement P(n): 3n < n! is true for all n ≥ 4 but P(1), P(2) and P(3) are not true.
View full question & answer→Question 195 Marks
Prove the following by the principle of mathematical induction:
$2.7^n + 3.5^{n-3}.3^{n-1}$^ is divisible of 24 for all $\text{n}\in\text{N}$
AnswerLet P(n) be the given statement.
Now,
$P(n): 2.7^n+ 3.5^n - 5$ is divisible by $24.$
Step 1:
$P(1): 2.7^1 + 3.5^1 - 5 = 24$
It is divisible by 24.
Thus, P(1) is true.
Step 2:
Let P(m) be true.
Then,
$2.7^m + 3.5^m - 5 $is divisible by $24.$
Suppose:
$2.7^\text{m}+3.5\text{m}-5=24\lambda \ ...(1)$
We need to show that P(m + 1) is true whenever P(m) is true.
Now,
$\text{P}(\text{m}+1)=2.7^{\text{m}+1}+3.5^{\text{m}+1}-5$
$=2.7^{\text{m}+1}+(24\lambda+5-2.7^\text{m})5-5$
$=(2.7^{\text{m}+1}+120\lambda+25-10.7^\text{m}5-5)$
$=2.7^{\text{m}}.7-10.7^\text{m}+120\lambda+24-4$
$=7^\text{m}(14-10)+120\lambda+24-4$
$=7^\text{m}.4+120\lambda+24-4$
$=4(7^\text{m}-1)+24(5\lambda+1)$
$\big[$Since $7^m - 1$ is a multiple of 6 for all $\text{n}\in\text{N},7^\text{m} - 1 = \mu\big]$
$=4\times6\mu+24(5\lambda+1)$
$=24(\mu+5\lambda+1)$
It is a multiple of 24.
Thus, P(m + 1) is true.
By the principle of mathematical induction, P(n) is true for $\text{n}\in\text{N}.$
View full question & answer→Question 205 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{n(n+1)}}=\frac{\text{n}}{\text{n}+1}$
AnswerLet P(n): $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{n(n+1)}}=\frac{\text{n}}{\text{n}+1}$
For n = 1
$\text{P}(1):\frac{1}{1.2}=\frac{1}{1+1}$
$\frac{1}{2}=\frac{1}{2}$
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{k(k+1)}}=\frac{\text{k}}{\text{k}+1}...(1)$
We have to show that
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{k(k+1)}}+\frac{\text{k}}{(\text{k}+1)(\text{k}+2)}=\frac{\text{k}+1}{(\text{k}+2)}$
Now,
$\Big\{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{k(k+1)}}\Big\}+\frac{\text{1}}{(\text{k}+1)(\text{k}+2)}$
$=\frac{\text{k}}{\text{k+1}}+\frac{1}{(\text{k+1)}(\text{k+2)}}$ [Using equation (1)]
$=\frac{1}{\text{k+1}}\Big[\frac{\text{k}(\text{k+2})+1}{(\text{k+2)}}\Big]$
$=\frac{1}{\text{k+1}}\Big[\frac{\text{k}^2+2\text{k}+1}{(\text{k+2)}}\Big]$
$=\frac{1}{\text{k+1}}\Big[\frac{(\text{k}+1)(\text{k}+2)}{(\text{k+2)}}\Big]$
$=\frac{(\text{k}+1)}{(\text{k+2)}}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 215 Marks
Prove that $\sin\text{x}+\sin3\text{x}+...+\sin(\text{2n-1})\text{x}=\frac{\sin ^2\text{nx}}{\sin\text{x}}$ for all $\text{n}\in\text{N}.$
AnswerLet P(n): $\sin\text{x}+\sin3\text{x}+...+\sin(\text{2n-1})\text{x}=\frac{\sin ^2\text{nx}}{\sin\text{x}}$
For n = 1
$\sin \text{x}=\frac{\sin^2\text{x}}{\sin\text{x}}$
$\sin\text{x}=\sin\text{x}$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, So
$\sin\text{x}+\sin3\text{x}+...+\sin(\text{2k-1})\text{x}=\frac{\sin ^2\text{kx}}{\sin\text{x}}$
We have to show that,
$\sin\text{x}+\sin3\text{x}+...+\sin(\text{2k-1})\text{x}+\sin(\text{2k+1})\text{x}=\frac{\sin ^2(\text{k+1})\text{x}}{\sin\text{x}}$
Now,
$\Big\{\sin\text{x}+\sin3\text{x}+...+\sin(\text{2k}-1)\text{x}\Big\}+\sin(2\text{k}+1)\text{x}$
$=\frac{\sin^2\text{kx}}{\sin\text{x}}+\frac{\sin(2\text{k}+1)\text{x}}{1}$
$=\frac{\sin^2\text{kx}+{\sin(2\text{k}+1)\text{x}}\sin\text{x}}{\sin\text{x}}$ [Using equation (1)]
$=\frac{2\sin^2\text{kx}+\cos\big[(2\text{k}+1)\text{x}-\text{x}-\big]\big[(2\text{k}+1)\text{x}+\text{x}\big]}{2\sin\text{x}}$
$=\frac{2\sin^2\text{kx}+\cos2\text{kx}-\cos(2\text{kx}+2\text{x})}{2\sin\text{x}}$
$=\frac{1-\cos2\text{kx}+\cos2\text{kx}-\cos2\text{x}(\text{k}+1)}{2\sin\text{x}}$
$=\frac{1-\cos2\text{x}(\text{k}+1)}{2\sin\text{x}}$
$=\frac{2\sin^2\text{x}(\text{k}+1)}{2\sin\text{x}}$
$=\frac{\sin^2\text{x}(\text{k}+1)}{\sin\text{x}}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 225 Marks
Prove the following by the principle of mathematical induction:
$2 + 5 + 8 + 11 + ... +(3\text{n} - 1) =\frac{1}{2}\text{n}(3\text{n}+1)$ $$
AnswerLet P(n): $2 + 5 + 8 + 11 + ... +(3\text{n} - 1) =\frac{1}{2}\text{n}(3\text{n}+1)$
for n = 1
$\text{P}(1)2=\frac{1}{2}.1.(4)$
2 = 2
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$2 + 5 + 8 + 11 + ... +(3\text{k} - 1) =\frac{1}{2}\text{k}(3\text{k}+1) \ ...(1)$
We have to show that
$2 + 5 + 8 + 11 + ... +(3\text{k} - 1)+(3\text{k} + 2) $
$=\frac{1}{2}(\text{k}+1)(3\text{k}+4) $
Now,
${2 + 5 + 8 + 11 + ... + (3\text{k} - 1)} + (3\text{k} + 2)$
$ =\frac{1}{2}\text{k}(3\text{k}+1)(3\text{k}+2)$
$=\frac{3\text{k}^2+\text{k}+2{(3\text{k}+2)}}{2}$
$=\frac{3\text{k}^2+\text{k}+6\text{k}+4}{2}$
$=\frac{3\text{k}^2+7\text{k}+4}{2}$
$=\frac{3\text{k}^2+3\text{k}+4\text{k}+4}{2}$
$\frac{(\text{k}+1)(3\text{k}+4)}{2}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N}$ by PMI.
View full question & answer→Question 235 Marks
Prove the following by the principle of mathematical induction:$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{n}-1)\text{d})=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
AnswerLet p(n): $\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{n}-1)\text{d})=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For n = 1
$\text{a}=\frac{1}{2}[2\text{a}+(1-1)\text{d}]$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, so
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{k}-1)\text{d})=\frac{\text{k}}{2}[2\text{a}+(\text{k}-1)\text{d}] \ ...(1)$
We have to show that
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{k}-1)\text{d})+(\text{a}+(\text{k})\text{d})=\frac{(\text{k+1})}{2}[2\text{a}+\text{k}\text{d}] $
Now,
$\big\{\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{k}-1)\text{d})\big\}+(\text{a}+\text{k}\text{d})$
$\frac{\text{k}}{2}[2\text{a}+(\text{k}-1)\text{d}]+(\text{a}+\text{k}\text{d})$ [Using equation (1)]
$=\frac{2\text{k}\text{a}+\text{k}(\text{k}-1)\text{d}+2(\text{a}+\text{k}\text{d})}{2}$
$=\frac{2\text{k}\text{a}+\text{k}^2\text{d}-\text{k}\text{d}+2\text{a}+2\text{k}\text{d})}{2}$
$=\frac{2\text{k}\text{a}+2\text{a}+\text{k}^2\text{d}+\text{k}\text{d}}{2}$
$=\frac{2\text{a}(\text{k}+1)+\text{d}(\text{k}^2 +\text{k})}{2}$
$=\frac{(\text{k}+1)}{2}[2\text{a}+\text{k}\text{d}]$
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 245 Marks
Prove the following by the principle of mathematical induction:
$1 + 2 + 2^2 + ... + 2^n = 2^{n+1} - 1$ for all $\text{n}\in\text{N}.$
AnswerLet p(n) be the statement given by
$p(n): 1 + 2 + 2^2 + ... + 2^n = 2^{n+1} - 1$
Step 1:
$p(1): 1 + 2^1 = 2^{1+1} - 1$
$\Rightarrow 1 + 2 = 4 - 1$
$\Rightarrow 3 = 3$
$\therefore$ p(1) is true.
Step 2:
Let p(m) is true. Then,
$1 + 2 + 2^2 + ... + 2^m = 2^{m+1} - 1 ... (1)$
We have to prove that P(m + 1) is true.
$1 + 2 + 2^2 + ... + 2^{m+1} = 1 + 2 + 2^2 + ... + 2^m + 2^{m+1}$
$= (2^{m+1} - 1) + 2^{m+1} ... [Using (1)]$
$= (2^{m+1} + 2^{m+1}) - 1$
$= 2 \times 2^{m+1} - 1$
$= 2^{m+2} - 1$
$\Rightarrow P(m + 1)$ is true.
Hence by the PMI,l the given result is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 255 Marks
Prove that $\frac{\text{n}^2}{7}+\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{\text{n}^2}{2}-\frac{37}{210}$ n is a positive integer for all $\text{n}\in\text{N}.$
AnswerLet p(n): $\frac{\text{n}^2}{7}+\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{\text{n}^2}{2}-\frac{37}{210}$ n is a positive integer
For n = 1
$\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}$
$=\frac{30+42+70+105-37}{210}$
$=\frac{247-37}{210}$
It is a positive integer
⇒ p(n) is true for n = 1
Let p(n) is true for n = k,
$\frac{\text{k}^2}{7}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{\text{k}^2}{2}-\frac{37}{210}$ k is positive integer
$\frac{\text{k}^2}{7}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{\text{k}^2}{2}-\frac{37}{210}\text{k}=\lambda$
For n = k + 1,
$\frac{(\text{k+1})^7}{7}+\frac{(\text{k+1})^5}{5}+\frac{(\text{k+1})^3}{3}+\frac{(\text{k+1})^2}{2}-\frac{37}{210}(\text{k+1})$
$=\frac{1}{7}\big[\text{k}^7+7\text{k}^6+21\text{k}^5+35\text{k}^3+21\text{k}^2+7\text{k}+1\big]\\+\frac{1}{5}\big[\text{k}^5+5\text{k}^4+10\text{k}^3+10\text{k}^2+21\text{k}^2+5\text{k}+1\big]\\+\frac{1}{3}\big[\text{k}^3+3\text{k}^2+3\text{k}+1\big]+\frac{1}{2}\big[\text{k}^2+2\text{k}+1\big]-\frac{37\text{k}}{210}-\frac{37}{210}$
$=\Big[\frac{\text{k}^2}{7}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{\text{k}^2}{2}-\frac{37}{210}\Big]+\Big[\text{k}^6+3\text{k}^5+5\text{k}^4+3\text{k}^2+\text{k}\\\ \ \ \ \ \ +\frac{1}{7}+\text{k}^4+2\text{k}^3+2\text{k}^2+\frac{1}{5}+\text{k}^2+\text{k}+\frac{1}{3}+\text{k}+\frac{1}{2}-\frac{37}{210}\Big]$
$=\lambda+\text{k}^6+3\text{k}^5+6\text{k}^4+7\text{k}^3+6\text{k}^2+3\text{k}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}-\frac{37}{210}$
$=\lambda+\text{k}^6+3\text{k}^5+6\text{k}^4+7\text{k}^3+6\text{k}^2+3\text{k}+1$
= Positive integer
p(n) is true for n = k + 1
p(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 265 Marks
Prove the following by the principle of mathematical induction:
$1.2 + 2.3 + 3.4 + ... +\text{n}(\text{n}+1)=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{3}$
AnswerLet P(n): $1.2 + 2.3 + 3.4 + ... +\text{n}(\text{n}+1)=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{3}$
For n = 1
$1.2=\frac{1(1+1)(1+2)}{3}$
2 = 2
⇒ P(n) is true for n = 1
Let P(n) is true for n = k
$\Rightarrow1.2 + 2.3 + 3.4 + ... +\text{k}(\text{k}+1)=\frac{\text{k}(\text{k}+1)(\text{k}+2)}{3} \ ...(1)$
We have to show that
$1.2 + 2.3 + 3.4 + ... +\text{k}(\text{k}+1)+(\text{k}+1)(\text{k}+2)$
$=\frac{(\text{k}+1)(\text{k}+2)(\text{k}+3)}{3}$
Now,
$\{1.2 + 2.3 + 3.4 + ... +\text{k}(\text{k}+1)\}+(\text{k}+1)(\text{k}+2)$
$=\frac{\text{k}(\text{k}+1)(\text{k}+2)}{3}+\frac{(\text{k}+1)(\text{k}+2)}{1}$
$=(\text{k}+1)(\text{k}+2)\Big[\frac{\text{k}}{3}+1\Big]$
$=\frac{(\text{k}+1)(\text{k}+2)(\text{k}+3)}{3}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 275 Marks
Prove that $\text{x}^{2\text{n}-1}+\text{y}^{2\text{n}-1}$ is divisible by x + y for all $\text{n}\in\text{N}.$
AnswerLet P(n): $\text{x}^{2\text{n}-1}+\text{y}^{2\text{n}-1}$ is divisible by (x + y)
For n = 1
$\text{x}^{2(\text{1})-1}+\text{y}^{2(\text{1})-1}$
= x + y
⇒ p(n) is true for n = 1
Let p(n) is true for n = k,
$\text{x}^{2\text{k}-1}+\text{y}^{2\text{k}-1}$ is divisible by (x + y)
$\text{x}^{2\text{k}-1}+\text{y}^{2\text{k}-1}=(\text{x + y})\lambda \ ...(1)$
We have to show that,
$\text{x}^{2\text{k}+1}+\text{y}^{2\text{k}+ 1}=(\text{x + y})\mu$
Now,
$\text{x}^{2\text{k}+1}+\text{y}^{2\text{k}+1}$
$\text{x}^{2\text{k}-1}.\text{x}^2+\text{y}^{2\text{k}+1}.\text{y}^2$
$=\Big[(\text{x}+\text{y})\lambda-\text{y}^{2\text{k}-1}\Big]\text{x}^2+\text{y}^{2\text{k}-1}.\text{y}^2$
$=(\text{x}+\text{y})\lambda\text{x}^2-\text{y}^{2\text{k}-1}.\text{x}^2+\text{y}^{2\text{k}-1}.\text{y}^2$
$=(\text{x}+\text{y})\lambda\text{x}^2-\text{y}^{2\text{k}-1}\big(\text{x}^2+\text{y}^2\big)$
$=(\text{x}+\text{y})\lambda\text{x}^2-\text{y}^{2\text{k}-1}\big(\text{x}+\text{y})(\text{x}+\text{y}\big)$
$=(\text{x}+\text{y})\Big[\lambda\text{x}^2-\text{y}^{2\text{k}-1}\big(\text{x}-\text{y}\big)\Big]$
$=(\text{x}+\text{y})\mu$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 285 Marks
Prove the following by the principle of mathematical induction:
$7^{2n} + 2^{3n-3}.3^{n-1}$ is divisible of 25 for all $\text{n}\in\text{N}$
AnswerLet $p(n): 7^{2n} + 2^{3n-3}.3^{n-1}$ is divisible of $25$
For $n = 1$
$7^2 + 2^0.3^0$
$= 49 + 1$
$= 50$
It is divisible of 25
$\Rightarrow p(n)$ is true for $n = 1$
Let p(n) is true for $n = k,$
$7^{2k} + 2^{3k-3}.3^{k-1}$ is divisible of $25$
$\Rightarrow7^{2\text{k}}+2^{3\text{k}-3}.3^{\text{k}-1}=25\lambda \ ...(1)$
We have to show that,
$7^{2(k+1)} + 2^{3k}3^k $ is divisible of $25$
$7^{2(\text{k}+1)}+2^{3\text{k}}.3\text{k}=25\mu$
Now,
$7^{2(\text{k}+1)}+2^{3\text{k}}.3\text{k}$
$=7^{2\text{k}}.7^2+2^{3\text{k}}.3^\text{K}$
$=(25\lambda-2^{3\text{k}-3}.3^{\text{k}-1})49+2^{3\text{k}}.3\text{k}$ [Using equation (1)]
$=25\lambda.49-\frac{2^{3\text{k}}}{8}.\frac{3^\text{k}}{3}.49+2^{3\text{k}}.3^\text{k}$
$=24.25.49\lambda-25.2^{3\text{k}}.3^\text{k}$
$25(24.49\lambda-2^{3\text{k}}.3^\text{k})$
$=25\mu$
$\Rightarrow p(n)$ is true for $n = k + 1$
$\Rightarrow p(n)$ is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 295 Marks
Prove that $\frac{\text{n}^{11}}{11}+\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{62}{165}$ n is a positive integer for all $\text{n}\in\text{N}.$
AnswerLet p(n): $\frac{\text{n}^{11}}{11}+\frac{\text{n}^5}{5}+\frac{\text{n}^3}{3}+\frac{62}{165}$ n is a positive integer
For n = 1
$\frac{1}{11}+\frac{1}{5}+\frac{1}{3}+\frac{62}{165}$
$=\frac{15+33+55+62}{165}$
$=\frac{165}{165}$
Which is a positive integer
Let p(n) is true for n = k, So
$\frac{\text{k}^{11}}{11}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{62}{165}$ is positive integer
$\frac{\text{k}^{11}}{11}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{62}{165}=\lambda \ ...(1)$
For n = k + 1,
$\frac{(\text{k+1})^{11}}{11}+\frac{(\text{k+1})^5}{5}+\frac{(\text{k+1})^3}{3}+\frac{62}{165}(\text{k+1})$
$=\frac{1}{11}\big[\text{k}^{11}+11\text{k}^{10}+55\text{k}^9+165\text{k}^8+330\text{k}^7+462\text{k}^6\\+462\text{k}^5+330\text{k}^4+165\text{k}^3+55\text{k}^2+11\text{k}+1\big]\\+\frac{1}{5}\big[\text{k}^5+5\text{k}^4+10\text{k}^3+10\text{k}^2+5\text{k}+1\big]\\+\frac{1}{3}\big[\text{k}^3+3\text{k}^2+3\text{k}+1\big]+\frac{62}{165}\big[\text{k}+1\big]$
$=\Big[\frac{\text{k}^{11}}{11}+\frac{\text{k}^5}{5}+\frac{\text{k}^3}{3}+\frac{62}{165}\Big]+\Big[\text{k}^{10}+5\text{k}^9+15\text{k}^8+30\text{k}^7+42\text{k}^6+42\text{k}^5+\\30\text{k}^4+15\text{k}^3+5\text{k}^2+1+\frac{1}{11}+\text{k}^4+2\text{k}^3+2\text{k}^2+\frac{1}{5}+\text{k}^2+\text{k}+\frac{1}{3}+\frac{62}{165}\Big]$
$=\lambda+\text{k}^{10}+5\text{k}^9+15\text{k}^8+30\text{k}^7+42\text{k}^6+42\text{k}^5+31\text{k}^4+17\text{k}^3+8\text{k}^3+2\text{k} +1$
$=\lambda+\text{k}^6+3\text{k}^5+6\text{k}^4+7\text{k}^3+6\text{k}^2+3\text{k}+1$
= An integer
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 305 Marks
Prove the following by the principle of mathematical induction:
$(ab)^n= a^nb^n f$ or all $\text{n}\in\text{N}$
AnswerLet $p(n): (ab)^n= a^nb^n$
For $n = 1$
$(ab)^1= a^1b^1$
$ab = ab$
$\Rightarrow p(n)$ is true for n = 1
Let p(n) is true for $n = k,$
$(ab)^k= a^kb^k ...(1)$
We have to show that,
$(ab)^{k+1}= a^{k+1}b^{k+1}$
Now,
$(ab)^{k+1}$
$= (ab)^k (ab)$
$= (a^kb^k)(ab)$ [Using equation $(1)]$
$= (a^{k+1})(b^{k+1})$
$\Rightarrow p(n)$ is true for $n = k + 1$
$\Rightarrow p(n)$ is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 315 Marks
$5^{2n} - 1$ is divisible by 24 for all $\text{n}\in\text{N}.$
AnswerLet $P(n)$ be the given statement.
Now,
$p(n): 5^{2 n}-1$ is divisible by 24 for all $n \in N$.
Step 1:
$p(1): 5^2-1=25-1=24$
It is divisible by 24 .
Thus, $p(1)$ is true.
Step 2:
Let $P(m)$ be true.
Then, $5^{2 m}-1$ is divisible by 24 .
Now, let $5^{2 m}-1=24 \lambda$, where $\lambda \in N$.
We need to show that $p(m+1)$ is true whenever $p(m)$ is true.
Now,
$P(m + 1) = 5^{2m+2} - 1$
$= 5^{2m}5^2 - 1$
$=25(24\lambda+1)-1$
$=600\lambda+24$
$=24(25\lambda+1)$
It is divisible by 24.
Thus, p(m + 1) is true.
By the principle of mathematical induction, p(n) is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 325 Marks
Prove the following by the principle of mathematical induction:
$\text{a}+\text{ar}+\text{ar}^2+...+\text{ar}^\text{n-1}=\text{a}\Big(\frac{\text{r}^\text{n}-1}{\text{r}-1}\Big),\text{r}\neq1$
AnswerLet p(n): $\text{a}+\text{ar}+\text{ar}^2+...+\text{ar}^\text{n-1}=\text{a}\Big(\frac{\text{r}^\text{n}-1}{\text{r}-1}\Big),\text{r}\neq1$
For n = 1
$\text{a}=\text{a}\Big(\frac{\text{r}^1-1}{\text{r}-1}\Big)$
a = a
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, so
$\text{a}+\text{ar}+\text{ar}^2+...+\text{ar}^\text{k-1}=\text{a}\Big(\frac{\text{r}^\text{k}-1}{\text{r}-1}\Big),\text{r}\neq1 \ ...(1)$
We have to show that
$\text{a}+\text{ar}+\text{ar}^2+...+\text{ar}^\text{k-1}=\text{a}\Big(\frac{\text{r}^\text{k+1}-1}{\text{r}-1}\Big)$
Now,
$\Big\{\text{a}+\text{ar}+\text{ar}^2+...+\text{ar}^\text{k-1}\Big\}+\text{ar}^\text{k}$
$=\text{a}\Big(\frac{\text{r}^\text{k}-1}{\text{r}-1}\Big)+\text{ar}^\text{k}$ [Using equation (1)]
$=\frac{\text{a}\big[\text{r}^\text{k}-1+\text{r}^\text{k}(\text{r}-1)\big]}{\text{r}-1}$
$=\frac{\text{a}\big[\text{r}^\text{k}-1+\text{r}^\text{k+1}-\text{r}^\text{k}\big]}{\text{r}-1}$
$=\text{a}\Big(\frac{\text{r}^\text{k+1}-1}{\text{r}-1}\Big)$
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 335 Marks
Prove the following by the principle of mathematical induction:
$1.2 + 2.2^2 + 3.2^3 + ... + n.2^n= (n - 1) 2^{n+1} + 2$
AnswerLet $P(n): 1.2 + 2.2^2 + 3.2^3 + ... + n.2^n= (n - 1) 2^{n+1} + 2$
for $n = 1$
$1.2 = 0.2^0 + 2$
$2 = 2$
$\Rightarrow P(n)$ is true for $n = k + 1$
$\Rightarrow P(n)$ is true for $n = k$, so
$1.2 + 2.2^2 + 3.2^3 + ... + k.2^k= (k - 1) 2^{k+1} + 2 ...(1)$
We have to show that
${1.2 + 2.2^2 + 3.2^3 + ... + k.2^k} + (k + 1) 2^{k+1} = k2^{k+2} + 2$
Now,
${1.2 + 2.2^2 + 3.2^3 + ... + k.2^k} + (k + 1) 2^{k+1}$
$= [(k - 1) 2^{k+1} + 2)] + (k + 1) 2^{k+1}$ [Using equation $(1)]$
$= (k - 1) 2^{k+1} + 2 + (k + 1) 2^{k+1}$
$= 2^{k+1} (k - 1 + k + 1) + 2$
$= 2^{k+1}.2k + 2$
$= k2^{k+2} + 2$
$\Rightarrow P(n)$ is true for $n = k + 1$
$\Rightarrow P(n)$ is true for all $\text{n}\in\text{N}$ by $PMI$
View full question & answer→Question 345 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^\text{n}}=1-\frac{1}{2^\text{n}}$
AnswerLet P(n): $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^\text{n}}=1-\frac{1}{2^\text{n}}$
For n = 1
$\frac{1}{2}=1-\frac{1}{2^1}$
$\frac{1}{2}=\frac{1}{2}$
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^\text{k}}=1-\frac{1}{2^\text{k}} \ ...(1)$
We have to show that
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^\text{k}}+\frac{1}{2^\text{k+1}}=1-\frac{1}{2^\text{k+1}}$
Now,
$\Big\{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^\text{k}}\Big\}+\frac{1}{2^\text{k+1}}$
$=1-\frac{1}{2^\text{k}}+\frac{1}{2^\text{k+1}}$ [Using equation (1)]
$=1-\Big(\frac{2-1}{2^\text{k+1}}\Big)$
$=1-\frac{1}{2^\text{k+1}}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 355 Marks
Let P(n) be the statement: $2^{\text{n}}\geq3\text{n}.$ If p(r) is true, Show that p(r + 1) is true. Do you conclude that p(n) is true for all $\text{n}\in\text{N}$?
AnswerP(n): $2^{\text{n}}\geq3\text{n}$
It is given that p(r) is true, So
$2^{\text{r}}\geq3\text{r} \ ...(1)$
Multiplying both the sides by 2,
$2^{\text{r}}.2\geq3\text{r}.2$
$2^{\text{r}+1}\geq6\text{r}$
$2^{\text{r}+1}\geq3\text{r}+3\text{r}$
$2^{\text{r}+1}\geq3+3\text{r} \ \big[\text{Since} \ 3\text{r}\geq3, \ 6\text{r}\geq3+3\text{r}\big]$
$2^{\text{r}+1}\geq3+3(\text{r} +1)$
So, p(r + 1)
But for r = 1
$2\geq3$
It is true, So
P(n) is not true for all $\text{n}\in\text{N}$ by PMI.
View full question & answer→Question 365 Marks
Using principle of mathematical induction prove that $\sqrt{\text{n}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{n}}}$ for all natural numbers $\text{n}\geq2.$
AnswerLet p(n) be the statement given by P(n):$\sqrt{\text{n}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{n}}}$ for all natural numbers $\text{n}\geq2.$Step 1:
p(2): $\sqrt{2}=1.4142$ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1+\frac{1}{1.4142}=1+0.7071=1.7071$ $\therefore\sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}$ $\therefore$ p(2) is true.Step 2:
Let p(m) is true. Then, $\sqrt{\text{m}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}} \ ...(1)$ We have to prove that p(m + 1) is true. $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}>\sqrt{\text{m}} \ ...[\text{From}(1)]$ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\sqrt{\text{m}}+\frac{1}{\sqrt{\text{m}+1}}$ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\frac{\sqrt{\text{m}^2+\text{m}}+1}{\sqrt{\text{m}+1}}$ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\frac{\sqrt{\text{m}^2}+1}{\sqrt{\text{m}+1}}$ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\frac{\text{m}+1}{\sqrt{\text{m}+1}}$ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{\text{m}}}+\frac{1}{\sqrt{\text{m}+1}}>\sqrt{\text{m}+1}$ ⇒ p(m + 1) is true. Hence by the principle of mathematical induction, the given results is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 375 Marks
Prove that $\stackrel{{7+77+777+...+777\ ...................\ 7}}{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \hbox{ n-digits}}$ $$${ =\frac{7}{81}(10^{\text{n}+1}-9\text{n}-10)}$ for all $\text{n}\in\text{N}.$
AnswerLet p(n) be the statement given by
p(n): $7+77+777+...+777\ ........\ 7 { =\frac{7}{81}(10^{\text{n}+1}-9\text{n}-10)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{n-digits}$ for all $\text{n}\in\text{N}.$
Step 1:
P(1) : $7-\frac{7}{81}[10^{1+1}-9(1)=10]$
$\Rightarrow7=\frac{7}{81}\times(100-9-10)$
$\Rightarrow7=\frac{7}{81}\times81$
$\Rightarrow7=7\times(1)$
$\therefore$ P(1) is true.
Step 2:
Let p(m) is true. Then,
$7+77+777+...+777\ ........\ 7 { =\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m-digits}$
We have to prove that p(m + 1) is true.
$7+77+777+...+777\ ........\ 7=7+77+777+...+777\ ........\ 7+777..............7 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m+1-digits} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m-digits}\ \ \ \ \ \ \ \ \ \ \ \ \text{m+1-digits}$
${ =\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)}+{7{[1111...............1]}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\text{m+1-digits}}$
${ =\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)}+{7{[1111...............1]}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m+1-digits}$
${ =\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)}+{7{[9999...............9]}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{m+1-digits}$
$=\frac{7}{81}(10^{\text{m}+1}-9\text{m}-10)+\frac{7}{9}[10^{\text{m}+1}-1]$
$=\frac{7}{81}\big[(1+9)10^{\text{m+1}}-9\text{m}-19\big]$
$=\frac{7}{81}\big[10\times10^{\text{m}+1}-9(\text{m+1)}-10\big]$
$=\frac{7}{81}\big[10\times10^{\text{m}+2}-9(\text{m+1)}-10\big]$
⇒ p(m + 1) is true.
View full question & answer→Question 385 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(\text{3n-1)(3n+2)}}=\frac{\text{n}}{\text{6n}+4}$
AnswerLet P(n): $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(\text{3n-1)(3n+2)}}=\frac{\text{n}}{\text{6n}+4}$For n = 1
$\text{P}(1):\frac{1}{2.5}=\frac{1}{6+4}$
$\frac{1}{10}=\frac{1}{10}$
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(\text{3k-1)(3k+2)}}=\frac{\text{k}}{\text{6k}+4} \ ... (1)$
We have to show that
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(\text{3k-1)(3k+2)}}+\frac{1}{(3\text{k}+2)(3\text{k}+5)}=\frac{(\text{k}+1)}{\text{6k}+10}$
Now,
$\Big\{\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(\text{3k-1)(3k+2)}}\Big\}+\frac{1}{(3\text{k}+2)(3\text{k}+5)}$
$=\frac{\text{k}(3\text{k}+5)+2}{2(3\text{k}+2)(3\text{k}+5)}$
$=\frac{3\text{k}^2+5\text{k}+2}{2(3\text{k}+2)(3\text{k}+5)}$
$=\frac{3\text{k}^2+3\text{k}+2\text{k}+2}{2(3\text{k}+2)(3\text{k}+5)}$
$=\frac{3\text{k}(\text{k}+1)+2(\text{k}+1)}{2(3\text{k}+2)(3\text{k}+5)}$
$=\frac{(\text{k}+1)+(3\text{k}+2)}{2(3\text{k}+2)(3\text{k}+5)}$
$=\frac{(\text{k}+1)}{2(3\text{k}+5)}$
View full question & answer→Question 395 Marks
A sequence $x_1, x_2, x_3, ...$ is defined by letting $x_1 = 2$ and $\text{x}_{\text{k}}=\frac{\text{x}_{\text{k}}-1}{\text{n}}$ for all natural numbers k, $\text{k}\geq2.$ Show that $\text{x}_{\text{n}}=\frac{2}{\text{n}!}$ for all $\text{n}\in\text{N}.$
AnswerLet p(n) be the statement given byP(n): $\text{x}_{\text{n}}=\frac{2}{\text{n}!}$ for all $\text{n}\in\text{N}.$
Step 1:
p(2): $\text{x}_{\text{n}}=\frac{2}{2!}=1$
Given that $\text{x}_{\text{k}}=\frac{\text{x}_{\text{k}}-1}{\text{n}}$ for all natural numbers $\text{k}\geq2$
$\text{x}_{2}=\frac{\text{x}_1}{2}=\frac{2}{2}=1$
$\therefore$ p(2) is true.
Step 2:
Let p(m) is true. Then,
$\text{x}_{\text{m}}=\frac{2}{\text{m}!} \ ...(1)$
We have to prove that p(m + 1) is true.
$\text{x}_{\text{m}+1}=\frac{\text{x}_{\text{m}+1-1}}{\text{m+1}}$
$\text{x}_{\text{m}+1}=\frac{\text{x}_{\text{m}}}{\text{m+1}}$
$\text{x}_{\text{m}+1}=\frac{{\frac{2}{\text{m}!}}}{\text{m+1}} \ ...[\text{From(1)}]$
$\text{x}_{\text{m}+1}=\frac{2}{\text{m}!(\text{m+1})}$
$\text{x}_{\text{m}+1}=\frac{2}{(\text{m+1})!}$
⇒ p(m + 1) is true.
Hence by the principle of mathematical induction, the given results is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 405 Marks
Prove the following by the principle of mathematical induction:
$1^2+3^2+5^2+...+(2\text{n}-1)^2=\frac{1}{3}\text{n}(4\text{n}^2-1)$
AnswerLet p(n): $1^2+3^2+5^2+...+(2\text{n}-1)^2=\frac{1}{3}\text{n}(4\text{n}^2-1)$
For n = 1
$1=\frac{1}{3}.1.(4-1)$
1 = 1
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, so
$1^2+3^2+5^2+...+(2\text{k}-1)^2=\frac{1}{3}\text{k}(4\text{k}^2-1) \ ...(1)$
We have to show that
$1^2+3^2+5^2+...+(2\text{k}-1)^2+(2\text{k}+1)^2=\frac{1}{3}(\text{k+1)}\big[(4(\text{k+1})^2-1)\big]$
Now,
$\big\{1^2+3^2+5^2+...+(2\text{k}-1)^2\big\}+(2\text{k}+1)^2$
$=\frac{1}{3}\text{k}(4\text{k}^2-1)+(2\text{k}-1)^2$ [Using equation (1)]
$=\frac{1}{3}\text{k}(2\text{k}+1)(2\text{k}-1)+(2\text{k}+1)^2$
$=(2\text{k}+1)\Big[\frac{\text{k}(2\text{k}-1)}{3}+(2\text{k}+1)\Big]$
$=(2\text{k}+1)\Big[\frac{2\text{k}^2-\text{k}+3(2\text{k}+1)}{3}\Big]$
$=(2\text{k}+1)\Big[\frac{2\text{k}^2-\text{k}+6\text{k}+3}{3}\Big]$
$=\frac{(2\text{k}+1)(2\text{k}^2+5\text{k}+3)}{3}$
$=\frac{(2\text{k}+1)(2\text{k}(\text{k}+1)+3(\text{k}+1))}{3}$
$=\frac{(2\text{k}+1)(2\text{k}+3)+(\text{k}+1)}{3}$
$=\frac{(\text{k}+1)}{2}\big[4\text{k}^2+6\text{k}+2\text{k}+3\big]$
$=\frac{(\text{k}+1)}{2}\big[4\text{k}^2+8\text{k}+4-1]$
$=\frac{(\text{k}+1)}{2}\big[4(\text{k}+1)^2-1]$
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 415 Marks
Prove the following by the principle of mathematical induction:
$n^3 - 7n + 3$ is divisible by 3 for all $\text{n}\in\text{N}$
AnswerLet p(n) be the statement given by
$p(n): n^3 - 7n + 3$ is divisible by $3.$
Step 1:
$p(1): 1^3 - 71 + 3$ is divisible by $3$
$\therefore 1 - 7 + 3 = -3$ is divisible by $3$
$\therefore p(1) $ is true
Step 2:
Let p(m) is true. Then,
$m^3 - 7m + 3$ is divisible by $3$
$\Rightarrow\text{m}^3-7\text{m}+3=3\lambda$ for some $\lambda\in\text{N} \ ...(1)$
We have tp prove that p(m + 1) is true.
$(m + 1)^3 - 7(m + 1) + 3 = m^3 + 3m^2+ 3m + 1 - 7m - 7 + 3$
$= m^3- 7m + 3 + 3m^2 + 3m + 1 - 7$
$= (m^3 - 7m + 3) + 3(m^2 + m - 2)$
$=3\lambda+3(\text{m}2+\text{m}-2)$ [Using equation (1)]
$=3[\lambda+(\text{m}^2+\text{m}-2)]$ which is divisible by 3
$\Rightarrow p(m + 1)$ is true.
Hence by the principle of mathematical induction, the given result is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 425 Marks
The distributive law from algebra states that for all real numbers $c, a_1$ and $a_2$, we have $c(a_1 + a_2)=ca_1 + ca_2.$
Use this law and mathematical induction to prove that, for all natural numbers $\text{n}\geq2,$ if $c, a_1, a_2, ...,$ an are any real numbers, then $c(a_1 + a_2 +...+ a_{n)}= ca_1 + ca_2 +...+ ca_n.$
AnswerThe distributive law from algebra states that for all real numbers c$, a_1$ and $a_2$, we have $c(a_1 + a_2) = ca_1 + ca_2$. Use this law and mathematical induction to prove that, for all natural numbers $\text{n}\geq2,$ if c,$ a_1, a_2, ...,$ an are any real numbers, then $c(a_1 + a_2 +...+ a_{n)}= ca_1 + ca_2 +...+ ca_n.$
Let p(n) be the statement given by
$P(n): c(a_1 + a_2 + ... + a_n) = ca_1 + ca_2 + ca_3 + ... + ca_n$ for all natural numbers $\text{n}\geq2.$
Step 1:
$p(2): c(a_1 + a_2) = ca_1 + ca_2$
$\therefore$ p(2) is true.
Step 2:
Let p(m) is true. Then,
$c(a_1 + a_2 + ... + a_m) = ca_1 + ca_2 + ca_3 + ... + ca_m ... (1)$
We have to prove that p(m + 1) is true.
$c(a_1 + a_2 + ... + a_m + a_{m+1}) = c[(a_1 + a_2 + ... + a_m) + a_{m+1}]$
$c(a_1 + a_2 + ... + a_m + a_{m+1}) = c(a_1 + a_2 + ... + a_m) + a_{m+1}$
$c(a_1 + a_2 + ... + a_m + a_{m+1}) = ca_1 + ca_2 + ca_3 ... + ca_m + ca_{m+1 [From (1)]}$
$\Rightarrow p(m + 1)$ is true.
Hence by the principle of mathematical induction, the given results is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 435 Marks
Given $\text{a}_1=\frac{1}{2}\Big(\text{a}_0+\frac{\text{A}}{\text{a}_0}\Big),\text{a}_2=\frac{1}{2}\Big(\text{a}_1+\frac{\text{A}}{\text{a}_1}\Big)$ and $\text{a}_{\text{n}+1}=\frac{1}{2}\Big(\text{a}_\text{n}+\frac{\text{A}}{\text{a}_\text{n}}\Big)$ for $\text{n}\geq2,$ Where a > 0, A > 0. Prove that $\frac{\text{a}_{\text{n}}-\sqrt{\text{A}}}{{\text{a}_{\text{n}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)^{2^{\text{n}-1}}$
Answer$\text{a}_1=\frac{1}{2}\Big(\text{a}_0+\frac{\text{A}}{\text{a}_0}\Big),\text{a}_2=\frac{1}{2}\Big(\text{a}_1+\frac{\text{A}}{\text{a}_1}\Big)$ and $\text{a}_{\text{n}+1}=\frac{1}{2}\Big(\text{a}_\text{n}+\frac{\text{A}}{\text{a}_\text{n}}\Big)$
Let P(n): $\frac{\text{a}_{\text{n}}-\sqrt{\text{A}}}{{\text{a}_{\text{n}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)^{2^{\text{n}-1}}$
For n = 1
$\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{{\text{a}_{\text{1}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)^{2^{\text{n}-1}}$
$\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{{\text{a}_{\text{1}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k,
$\frac{\text{a}_{\text{k}}-\sqrt{\text{A}}}{{\text{a}_{\text{k}}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg) \ ...(1)$
We have to show that,
$\frac{\text{a}_{\text{k}+1}-\sqrt{\text{A}}}{{\text{a}_{\text{k}+1}+\sqrt{\text{A}}}}=\Bigg(\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{\text{a}_{\text{1}}+\sqrt{\text{A}}}\Bigg)^{2\text{k}}$
$\Bigg(\frac{\text{a}_{\text{k}+1}-\sqrt{\text{A}}}{{\text{a}_{\text{k}+1}+\sqrt{\text{A}}}}\Bigg)^{2^0}$
$=\begin{bmatrix}\frac{\frac{\frac{1}{2}\Big(\text{a}_\text{k}+\frac{\text{A}}{\text{a}_\text{k}}\Big)-\sqrt{\text{A}}}{1}}{2\Big(\text{a}_\text{k}+\frac{\text{A}}{\text{a}_\text{k}}\Big)+\sqrt{\text{A}}} \end{bmatrix}^{2^0}$
$=\Bigg[\frac{(\text{a}_\text{k})^2+\text{A}-2\text{a}_\text{k}\sqrt{\text{A}}}{(\text{a}_\text{k})^2+\text{A}+2\text{a}_\text{k}\sqrt{\text{A}}}\Bigg]^{2^0}$
$=\frac{\Big(\text{a}_{\text{k}}-\sqrt{\text{A}}\Big)^2}{\Big({\text{a}_{\text{k}}+\sqrt{\text{A}}\Big)^2}}$
$=\Bigg[\frac{\text{a}_{\text{k}}-\sqrt{\text{A}}}{{\text{a}_{\text{k}}+\sqrt{\text{A}}}}\Bigg]^{2^1}$
$=\Bigg[\frac{\text{a}_{\text{1}}-\sqrt{\text{A}}}{{\text{a}_{\text{1}}+\sqrt{\text{A}}}}\Bigg]^{2^\text{k}}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PM
View full question & answer→Question 445 Marks
Prove that $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{\text{n}^2}<2-\frac{1}{\text{n}}$ for all $\text{n}>2,$ $\text{n}\in\text{N}.$
AnswerLet P(n): $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{\text{n}^2}<2-\frac{1}{\text{n}}$ for all $\text{n}>2$
For n = 2
$1+\frac{1}{4}<2-\frac{1}{4}$
$=\frac{5}{4}<\frac{7}{4}$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, So
$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{\text{k}^2}<2-\frac{1}{\text{k}} \ ...(1)$
We have to show that,
$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{\text{k}^2}+\frac{1}{(\text{k+1})^2}<2-\frac{1}{(\text{k+1})}$
Now,
$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{\text{k}^2}+\frac{1}{(\text{k+1})^2}$
$<2-\frac{1}{\text{k}}+\frac{1}{(\text{k+1})^2}$ [Using equation (1)]
$<2-\frac{\text{k}^2+2\text{k}+1-\text{k}}{\text{k}(\text{k+1})^2}$
$<2-\frac{\text{k}^2+\text{k}+1}{\text{k}(\text{k+1})^2}$
$<2-\frac{\text{k}^2+\text{k}}{\text{k}(\text{k+1})^2}$
$<2-\frac{\text{k}(\text{k}+1)}{\text{k}(\text{k+1})^2}$
$<2-\frac{1}{(\text{k+1})}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 455 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(\text{2n+1)(2n+3)}}=\frac{\text{n}}{3(\text{2n}+1)}$
AnswerLet P(n): $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(\text{2n+1)(2n+3)}}=\frac{\text{n}}{3(\text{2n}+1)}$
For n = 1
$\text{P}(1):\frac{1}{3.5}=\frac{1}{15}$
$\frac{1}{15}=\frac{1}{15}$
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(\text{2k+1)(2k+3)}}=\frac{\text{k}}{3(\text{2k}+1)} \ ...(1)$
We have to show that
$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(\text{2k+1)(2k+3)}}+\frac{1}{(2\text{k}+3)(2\text{k}+5)}=\frac{(\text{k}+1)}{3\text{(2k}+5)}$
Now,
$\Big\{\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(\text{2k+1)(2k+3)}}\Big\}+\frac{1}{(2\text{k}+3)(2\text{k}+5)}$
$=\frac{\text{k}}{3(2\text{k}+3)}+\frac{1}{(2\text{k}+3)(2\text{k}+5)}$ [Using equation (1)]
$=\frac{1}{(2\text{k}+3)}+\Big[\frac{\text{k}}{3}+\frac{1}{(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}+\Big[\frac{\text{k}(2\text{k}+5)+3}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}+\Big[\frac{2\text{k}^2+5\text{k}+3}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}+\Big[\frac{(2\text{k}^2+2\text{k}+3\text{k}+3)}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}+\Big[\frac{(2\text{k}(\text{k}+1)+3(\text{k}+1)}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}+\Big[\frac{(2\text{k}+3)(\text{k}+1)}{3(2\text{k}+5)}\Big]$
$=\frac{(\text{k}+1)}{3(2\text{k}+5)}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N} $ by PMI
View full question & answer→Question 465 Marks
Prove that $\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{n}}}\tan\Big(\frac{\text{x}}{2^{\text{n}}}\Big)=\frac{1}{2^{\text{n}}}\cot\Big(\frac{\text{x}}{2^{\text{n}}}\Big)-\cot\text{x}$ for all $\text{n}\in\text{N}$ and $0<\text{x}<\frac{\pi}{2}$
AnswerLet p(n): $\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{n}}}\tan\Big(\frac{\text{x}}{2^{\text{n}}}\Big)=\frac{1}{2^{\text{n}}}\cot\Big(\frac{\text{x}}{2^{\text{n}}}\Big)-\cot\text{x}$
For n = 1
$\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)=\frac{1}{2}\cot\Big(\frac{\text{x}}{2}\Big)-\cot\text{x}$
$=\frac{1}{2}\frac{1}{2\tan\frac{\text{x}}{2}}-\frac{1}{\tan\text{x}}$
$=\frac{1}{2\tan\frac{\text{x}}{2}}-\frac{1}{\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1-2\tan^2\frac{\text{x}}{2}}\Bigg)}$
$=\frac{1}{2\tan\frac{\text{x}}{2}}-\frac{1-\tan^2\frac{\text{x}}{2}}{2\tan\frac{\text{x}}{2}}$
$=\frac{1-1+\tan^2\frac{\text{x}}{2}}{2\tan\frac{\text{x}}{2}}$
$=\frac{\tan^2\frac{\text{x}}{2}}{2\tan\frac{\text{x}}{2}}$
$=\frac{1}{2}\tan\frac{\text{x}}{2}$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, So
$\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{k}}}\tan\Big(\frac{\text{x}}{2^{\text{k}}}\Big)=\frac{1}{2^{\text{k}}}\cot\Big(\frac{\text{x}}{2^{\text{k}}}\Big)-\cot\text{x}$
We have to show that,
$\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{k}}}\tan\Big(\frac{\text{x}}{2^{\text{k}}}\Big)+\frac{1}{2^{\text{k+ 1}}}\tan\Big(\frac{\text{x}}{2^{\text{k+1}}}\Big)=\frac{1}{2^{\text{k+1}}}\cot\Big(\frac{\text{x}}{2^{\text{k+1}}}\Big)-\cot\text{x}$
Now,
$\Bigg\{\frac{1}{2}\tan\frac{\text{x}}{2}+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{k}}}\tan\Big(\frac{\text{x}}{2^{\text{k}}}\Big)\Bigg\}+\frac{1}{2^{\text{k+ 1}}}\tan\Big(\frac{\text{x}}{2^{\text{k+1}}}\Big)$
$=\frac{1}{2^\text{k}}\cot\Big(\frac{\text{x}}{2^\text{k}}\Big)-\cot\text{x}+\frac{1}{2^{\text{k+1}}}\tan\Big(\frac{\text{x}}{2^{\text{k}+1}}\Big)$
$=\frac{1}{2^\text{k}}\cot\Big(\frac{\text{x}}{2^\text{k}}\Big)-\cot\text{x}+\frac{1}{2.2^{\text{k}}}\frac{1}{\cot\Big(\frac{\text{x}}{2^\text{k}}.\frac{1}{2}\Big)}$
$=\frac{1}{2^\text{k}}\Bigg[\frac{1}{\tan\Big(\frac{\text{x}}{2^\text{k}}\Big)}+\frac{1}{2}.\tan\bigg\{\Big(\frac{\text{x}}{2^\text{k}}\Big).\frac{1}{2}\bigg\}\Bigg]-\cot\text{x}$
$=\frac{1}{2^\text{k}}\Bigg[\frac{1-\tan^2\Big(\frac{\text{x}}{2^\text{k+1}}\Big)}{2\tan\Big(\frac{\text{x}}{2^\text{k+1}}\Big)}+\frac{1}{2}\tan\Big(\frac{\text{x}}{2.2^\text{k}}\Big)\Bigg]-\cot\text{x} $
$=\frac{1}{2^\text{k}}\Bigg[\frac{1-\tan^2\Big(\frac{\text{x}}{2^{\text{k}+1}}\Big)+\tan^2\Big(\frac{\text{x}}{2^\text{k+1}}\Big)}{2\tan\Big(\frac{\text{x}}{2^\text{k+1}}\Big)}\Bigg]-\cot\text{x}$
$=\frac{1}{2^{\text{k}+1}}\Bigg[\frac{1}{\tan\Big(\frac{\text{x}}{2^{\text{k}+1}}\Big)}\Bigg]-\cot\text{x}$
$=\frac{1}{2^{\text{k}+1}}\cot\Big(\frac{\text{x}}{2^{\text{k}+1}}\Big)-\cot\text{x}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI
View full question & answer→Question 475 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(\text{3n-2)(3n+1)}}=\frac{\text{n}}{\text{3n}+1}$
AnswerLet P(n): $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(\text{3n-2)(3n+1)}}=\frac{\text{n}}{\text{3n}+1}$
For n = 1
$\text{P}(1):\frac{1}{1.4}=\frac{1}{4}$
$\frac{1}{4}=\frac{1}{4}$
⇒ P(n) is true for n = 1
Let P(n) is true for n = k, so
$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(\text{3k-2)(3k+1)}}=\frac{\text{k}}{\text{3k}+1} \ ...(1)$
We have to show that
$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(\text{3k-2)(3k+1)}}+\frac{1}{(3\text{k}+1)(3\text{k}+4)}=\frac{(\text{k}+1)}{\text{(3k}+4)}$
Now,
$\Big\{\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(\text{3k-2)(3k+1)}}\Big\}+\frac{1}{(3\text{k}+1)(3\text{k}+4)}$
$=\frac{\text{k}}{(3\text{k}+1)}+\frac{1}{(3\text{k}+1)(3\text{k}+4)}$
$=\frac{\text{k}}{(3\text{k}+1)}+\Big[\frac{\text{k}}{1}+\frac{1}{(3\text{k}+4)}\Big]$
$=\frac{1}{(3\text{k}+1)}+\Big[\frac{\text{k}(3\text{k}+4)+1}{(3\text{k}+4)}\Big]$
$=\frac{1}{(3\text{k}+1)}+\Big[\frac{3\text{k}^2+4\text{k}+1}{(3\text{k}+4)}\Big]$
$=\frac{1}{(3\text{k}+1)}+\Big[\frac{(3\text{k}^2+3\text{k}+\text{k}+1)}{(3\text{k}+4)}\Big]$
$=\frac{3\text{k}(\text{k}+1)+(\text{k}+1)}{(3\text{k}+1)(3\text{k}+4)}$
$=\frac{(\text{k}+1)+(3\text{k}+1)}{(3\text{k}+1)(3\text{k}+4)}$
$=\frac{(\text{k}+1)}{(3\text{k}+4)}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all $\text{n}\in\text{N} $ by PMI
View full question & answer→Question 485 Marks
Prove the following by the principle of mathematical induction:
$3^{2n+2} + 8n - 9$ is divisible by 8 for all $\text{n}\in\text{N}$
AnswerLet $p(n):3^{2n+2} + 8n - 9$ is divisible by $8$
For $n = 1$
$3^{2+2} - 8 - 9$
$=81- 17$
$= 64$
It is divisible by 8
$\Rightarrow p(n)$ is true for $n = 1$
Let p(n) is true for n = k, so
$(3^{2k+2} - 8k - 9)$ is divisible by $8$
$\Rightarrow3^{2\text{k}+2} - 8\text{k} - 9=8\lambda \ ...(1)$
We have to show that,
$3^{2(k+1)+2} - 8(k+1) - 9$ is divisible by $8$
$3^{2(\text{k}+1)}.3^2 - 8(\text{k}+1) - 9=8\mu$
Now,
$3^{2(\text{k}+1)}.9 - 8\text{k} - 8 - 9$
$=(8\lambda+8\text{k}+9)9-8\text{k}-8-9$
$=72\lambda+72\text{k}+81-8\text{k}-17$
$=72\lambda+64\text{k}+64$
$=8(9\lambda+8\text{k}+8)$
$=8\mu$
$\Rightarrow p(n)$ is true for $n = k + 1$
$\Rightarrow p(n)$ is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 495 Marks
A sequence $x_0, x_1, x_2, x_3, \ldots$ is defined by letting $x_0=5$ and $x_k=4+x_{k-1}$ for all natural numbers $k$. Show that $x_n=5+$ 4 n for all $n \in N$ using mathametical induction.
AnswerLet p(n) be the statement given by $P(n): x_n = 5 + 4n$ for all $\text{n}\in\text{N}.$
Step 1:
$p(1): x_1 = 5 + 4(1) = 5$ Given that $x_k = 5 + 4k$ for all natural numbers $k x_1 = 4 + x_0+ 5 = 9$ $\therefore$ p(1) is true.
Step 2:
$x_m = 5 + 4m ... (1)$ Let p(m) is true.
Then, We have to prove that $p(m + 1)$ is true. $x_{m+1} = 4 + x_{m+1-1} x_{m+1} = 4 + x_m x_{m+1} = 4 + 5 x_m ... [from (1)] x_{m+1} = 5 + 4(m + 1) \Rightarrow p(m + 1) $is true.
Hence by the principle of mathematical induction, the given results is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 505 Marks
Prove the following by the principle of mathematical induction:
n(n + 1)(n + 5) is a multiple of 3 for all $\text{n}\in\text{N}$
AnswerLet p(n): n(n + 1)(n + 5) is a multiple of 3 for all $\text{n}\in\text{N}$
For n = 1
1.(1 + 1)(1 + 5)
= (2)(6)
= 12
It is a multiple of 3
Let p(n) is true for n = k
k(k + 1)(k + 5) is a multiple of 3
$\text{k}(\text{k} + 1)(\text{k} + 5)=3\lambda \ ...(1)$
we have to show that,
(k + 1)(k + 2)[(k + 1)+5]
= [k(k + 1)+2(k + 1)][(k + 5)+1]
= k(k + 1)(k + 5) + k(k + 1) + 2k(k + 1)(k + 5) + 2(k + 1)
$=3\lambda+\text{k}^2+\text{k}+2(\text{k}^2+6\text{k}+5)+2\text{k}+2$ [Using equation (1)]
$=3\lambda+\text{k}^2+\text{k}+2\text{k}^2+12\text{k}+10+2\text{k}+2$
$=3\lambda+3\text{k}^2+15\text{k}+12$
$3(\lambda+\text{k}^2+5\text{k}+4)$
$=3\mu$
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI
View full question & answer→Question 515 Marks
Show by the Principle of Mathematical induction that the sum $S_n$ of the n terms of the series $1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + ...$ is given by
$\text{S}_\text{n}=\begin{cases}\frac{\text{n}(\text{n}+1)^2}{2},\text{if n is even}\\\frac{\text{n}^2(\text{n}+1)}{2},\text{if n is odd}\end{cases}$
Answer$S_n= 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + ...$
Using induction we first show this is true for n = 2,
We get $S_2 = 1^2 + 2 \times 2^2 = 1 + 8 = 9$
From RHS, we have if n is even $\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)^2}{2}$
$\text{S}_\text{2}=\frac{2\times9}{2}=9$
Now using induction we first shiw this is true also
for n = 3, we get $S_3 = 1 + 8 + 9 = 18$
From RHS, we have if n is odd $\text{S}_\text{n}=\frac{\text{n}^2(\text{n}+1)}{2}$
$\text{S}_\text{3}=\frac{9\times4}{2}=18$
Lets assume above is true for n = k, we get
$K$ is even, $S_k = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + ... + 2 \times k^2 ... (1)$
$K$ is odd, $S_k = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + ... + k^2 ... (2)$
Now lets prove for $n = k + 1$
If k is even, k + 1 is odd we get
$S_{k+1} = 1^2 + 2 \times 2^2 + 3^2 + ... + 2 \times k^2 + (k + 1)^2 ... (3)$
From above relation, we get
$S_k = 1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + ... + 2 \times k^2$ $=\frac{\text{k}(\text{k}+1)^2}{2}$
Substitute this on 3, we get
$\text{S}_{\text{k+1}}=\frac{\text{k}(\text{k}+1)^2}{2}+(\text{k}+1)^2=\frac{(\text{k}+1)^2(\text{k}+2)}{2}$
= RHS (when 'k + 1' is odd)
Hence Proved
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