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Maths Solve the Following Question.(4 Marks)

Mathematical Induction · Maharashtra Board STD 11 (MSBSHSE - Semi English Medium). 22 questions.

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Solve the Following Question.(4 Marks)

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Question 14 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{(\text{4n-1)(4n+3)}}=\frac{\text{n}}{3(\text{4n}+3)}$
Answer
Let 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
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Question 24 Marks
Prove the following by the principle of mathematical induction:
$1 + 3 + 3^2 + ... + 3^{\text{n}-1}=\frac{3^\text{n}-1}{2}$
Answer
Let $P(n)$ 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 $\text{n}\in\text{N}.$
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Question 34 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 $\text{n}\in\text{N}.$
Answer
Let
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.
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Question 44 Marks
Prove the following by the principle of mathematical induction:
$1.3 + 3.5 + 5.7 + ... + (2\text{n} - 1)(2\text{n} + 1)= \frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$
Answer
Let 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
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Question 54 Marks
Prove that $\frac{(2\text{n})!}{2^{2\text{n}}(\text{n}!)^2}\leq\frac{1}{\sqrt{3\text{n}+1}}$ for all $\text{n}\in\text{N}.$
Answer
P(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
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Question 64 Marks
Prove the following by the principle of mathematical induction:
1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! - 1 for all $\text{n}\in\text{N}$
Answer
Consider 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
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Question 74 Marks
A sequence $a_1, a_2, a_3, ...$ is defined by letting $a_1 = 3$ and $a_k = 7a_k - 1$ for all natural numbers $\text{k}\geq2.$ Show that $a_n = 3.7^{n-1}$ for all^$\text{n}\in\text{N}.$
Answer
Let $p(n)$ 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 $\text{n}\in\text{N}.$
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Question 84 Marks
Prove the following by the principle of mathematical induction:
$5^{2 n+2}+24 n-25$ is divisible by 576 for all $n \in N$
Answer

Let $p(n): 5^{2 n+2}+24 n-25$ 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
$5^{2 k+2}-24 k -25$ is divisible by 576
$5^{2 k+2}-24 k-25=576 \lambda \ldots(1)$
We have to show that,
$5(2 k+2)+2-24(k+1)-25$ 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)]
$=25.576\lambda+600\text{k}+625-24\text{k}-49$
$=25.576\lambda+576\text{k}+576$
$=576(25\lambda+\text{k}+1)$
$=576\mu$
$\Rightarrow p(n)$ is true for $n = k + 1$
$\Rightarrow p(n)$ is true for all by $\text{n}\in\text{N}$ PMI
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Question 94 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 $n^2​​​​​​​$​​​​​​​.
Answer
Let $P(n): 1+3+5+\ldots+(2 n-1)=n^2$
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,
$\{1+3+5+\ldots+(2 k-1)\}+2(k+1)$
$=k^2+2(k+1)[\text { Using equation (1)] }$
$=k^2+2 k+1$
$=(k+1)^2$
$\Rightarrow P(n) \text { is true for } n=k+1$
$\Rightarrow P ( n )$ is true for all $n \in N$ by PMI
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Question 104 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:
$\text{S}_\text{k+1}=\frac{1}{\text{k}+2}+\frac{1}{\text{k}+3}+...+\frac{1}{2\text{k}+2}$
$=\frac{-1}{\text{k}+1}+\frac{1}{\text{k}+1}+\frac{1}{\text{k}+2}+\frac{1}{\text{k}+3}+...+\frac{1}{2\text{k}}+\frac{1}{2\text{k}+1}+\frac{1}{2\text{k}+2}$
$=\frac{-1}{\text{k}+1}+\text{S}_\text{k}+\frac{1}{2\text{k}+1}+\frac{1}{2\text{k}+2}$
$=\text{S}_\text{k}+\frac{1}{2(2\text{k}+1)(\text{k}+1)}$
$>\text{S}_\text{k}$
$\therefore\text{S}_\text{k+1}>\frac{13}{24}$
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Question 114 Marks
Prove the following by the principle of mathematical induction:
$3^{2n} + 7$ is divisible by 8 for all $\text{n}\in\text{N}$
Answer
Let $p ( n ): 3^{2 n }+7$ is divisible by 8 for all $n \in N$
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
$3^{2 k}+7$ is divisible by 8
$\Rightarrow 3^{2 k}+7=8 \lambda \ldots(1)$
We have to show that,
$3^{2(k+1)}+7$ is divisible by 8
$3^{(2 k +1)}+7=8 \mu$
Now,
$3^{2(k+1)}+7$
$=3^{2 k} \cdot 3^2+7$
$=9.3^{2 k}+7$
$=9 .(8 \lambda-7)+7$
$=72 \lambda-56$
$=8(9 \lambda-7)$
$=8 \mu$
$\Rightarrow p(n)$ is true for $n=k+1$
$\Rightarrow p(n)$ is true for all by $n \in N$ PMI
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Question 124 Marks
Prove the following by the principle of mathematical induction:
$1.3 + 2.4 + 3.5 + ... + \text{n}(\text{n} + 2) $
$=\frac{1}{6}\text{n}(\text{n}+1)(2\text{n}+7)$
Answer
Let $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
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Question 134 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
$=\frac{\text{n}+1}{2\text{n}}$
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Question 144 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 $\frac{\text{n}(\text{n}+1)}{2}.$
Answer
Let 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 $\text{n}\in\text{N}$
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Question 154 Marks
Prove the following by the principle of mathematical induction:
$11^{n+2} + 12^2^{n+1}$ is divisible of 133 for all $\text{n}\in\text{N}$
Answer
Let $p ( n ): 11^{ n +2}+12^{2 n +1}$ 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
$11^{ k +2}+12^{2 k +1}=133 \lambda \ldots(1)$
We have to show that, $1^{k+3}+12^{2 k+3}$ is divisible of 133
Now,
$11^{ k +2} \cdot 11+12^{2 k+1} \cdot 12^2$
$=\left(133 \lambda-12^{2 k +1}\right) 11+12^{2 k +1} \cdot 144$
$=11 \cdot 133 \lambda+11 \cdot 12^{2 k+1}+144 \cdot 12^{2 k +1}$
$=11 \cdot 133 \lambda+133 \cdot 12^{2 k+1}$
$=133\left(11 \lambda+12^{2 k +1}\right)$
$=133 \mu$
$\Rightarrow p(n) \text { is true for } n=k+1$
$\Rightarrow p(n) \text { is true for all by } n \in N \text { PMI }$
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Question 164 Marks
Prove the following by the principle of mathematical induction:
$1^2 + 2^2 + 3^2 +...+ \text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
Answer
Let $P(n): 1^2+2^2+3^2+\ldots+n^2=\frac{n(n+1)(2 n+1)}{6}$
For $n =1$
$P(1): 1=\frac{1(1+1)(2+1)}{6}$
$1=1$
$\Rightarrow P(n)$ is true for $n=1$
Let $P(n)$ is true for $n=k$, so
$P(k): 1^2+2^2+3^2+\ldots+k^2=\frac{k( k +1)(2 k+1)}{6} \cdots$
We have to show that $P(n)$ is true for $n=k+1$
$\Rightarrow 1^2+2^2+3^2+\ldots .+k^2+(k+1)^2=\frac{( k +1)( k +2)(2 k +3)}{6}$
So, $1^2+2^2+3^2+\ldots . .+k^2+(k+1)^2$
$=\frac{ k ( k +1)(2 k +1)}{6}+( k +1)^2$ [Using equation (1)]
$=(k+1)\left[\frac{2 k^2+k}{6}+\frac{(k+1)}{1}\right]$
$=( k +1)\left[\frac{2 k ^2+ k +6 k +6}{6}\right]$
$=( k +1)\left[\frac{2 k ^2+7 k +6}{6}\right]$
$=(k+1)\left[\frac{2 k ^2+4 k +3 k +6}{6}\right]$
$=(k+1)\left[\frac{2 k(k+2)+3(k+2)}{6}\right]$
$=\frac{(k+1)(2 k+3)(k+2)}{6}$
$\Rightarrow P(n)$ is true for $n=k+1$
$\Rightarrow P ( n )$ is true for all $n \in N$ by PMI
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Question 174 Marks
Prove that the number of subsets of a set containing n distinct elements is $2^n$​​​​​​​ for all $\text{n}\in\text{N}.$
Answer
Let $p(n)$ 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 .
$\therefore P(1) \text { is true. }$
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 $n \in N$.
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Question 184 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.
Answer
Let 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.
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Question 194 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 $\text{n}\in\text{N}$
Answer
Let $P(n)$ 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 $\text{n}\in\text{N}.$
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Question 204 Marks
Prove the following by the principle of mathematical induction:
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\text{n(n+1)}}=\frac{\text{n}}{\text{n}+1}$
Answer
Let 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
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Question 214 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 $\text{n}\in\text{N}.$
Answer
Let 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
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Question 224 Marks
Prove the following by the principle of mathematical induction:
$2 + 5 + 8 + 11 + ... +(3\text{n} - 1) =\frac{1}{2}\text{n}(3\text{n}+1)$ $$
Answer
Let 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.
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