Question 515 Marks
Prove the following statement by principle of mathematical induction:
For any natural number n, $x^n - y^n$ is divisible by $x - y,$ where $x$ and $y$ are any integers with $x \neq y.$
Answer$P(n): x^n-y^n$ is divisible by $x-y$, where $x$ and $y$ are any integers with $x \neq y$.
Now, $P(1)$ : $x^1-y^1=x-y$, which is divisible by $(x-y)$
Hence, $P ( l )$ is true.
Let us assume that $P(n)$ is true for some natural number $n=k$.
$P(k): x^k-y^k$ is divisible by $(x-y)$
Or $x^k-y^k=m(x-y), m \neq N \ldots...(i)$
Now, we have to prove that P(k + 1) is true.
$P(k + 1): x^{k + l} - y^{k + l}$
$= x^k - x - x^k - y + x^k - y - y^ky$
$= x^k(x - y) + y(x^k - y^k)$
$= x^k(x - y) + ym(x - y) (Using (i))$
$= (x - y)[x^k + ym$], which is divisible by $(x - y)$
Hence, P(k + 1) is true whenever P(k) is true.
So,by the principle of mathematical induction P(n) is true for any natural number n.
View full question & answer→Question 525 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 535 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 545 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
⇒ 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 555 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 565 Marks
Prove the following statement by principle of mathematical induction:
$n^2< 2^n$ for all natural numbers $\text{n}\geq5$
AnswerLet $P(n): n^2 < 2^n$ for all natural number, $\text{n}\geq5$
Step 1:$ P(5): 5^2 < 2^5 \Rightarrow 25 < 32$ which is true. So P(5) is true.
Step 2: Assume P(k) is true for some $\text{k }\in\text{ N},\text{ k}\geq5\Rightarrow\text{P(k): }\text{k}^2<2^{\text{k}}\ ....(\text{i})$ is true.
Step 3: Now we have to prove $P(k + 1): (k + 1)^2 < 2^{k+1}$ is true where $\text{k}\geq5$
Now we have $(k + 1)^2= k^2+ 2k + 1 < 2^k + 2k + 1 .....(ii)$
Now let $2^k + 2k + 1 < 2^{k+1}$
$\Rightarrow 2^k + 2k + 1 < 2^k. 2 = 2^k + 2^k$
$\Rightarrow (2k + 1) < 2^k$ which is true for $\text{k }\in\text{ N},\text{ k}\geq5\ ...(\text{iii})$
So, $(k + 1)^2 < 2^k + 2^k$ [Using (ii) and (iii)]
$\Rightarrow (k + 1)^2 < 2.2^k$
$\Rightarrow (k + 1)^2 < 2^{k+1}$
$\Rightarrow P(k + 1)$ is true.
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have P(n) is true for all n.
View full question & answer→Question 575 Marks
Prove the following by the principle of mathematical induction:
$(ab)^n= a^nb^n$ for 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 585 Marks
$5^{2n} - 1$ is divisible by 24 for all $\text{n}\in\text{N}.$
AnswerLet P(n) be the given statement.
Now,
$p(n): 5^{2n} - 1$ is divisible by 24 for all $\text{n}\in\text{N}.$
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, $5^{2m} - 1$ is divisible by $24.$
Now, let $5^{2\text{m}}-1=24\lambda,$ where $\lambda\in\text{N}.$
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 595 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 605 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 615 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 625 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 635 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 645 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 655 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 665 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 675 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 685 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
⇒ 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 695 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 705 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 715 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 725 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 735 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 745 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 755 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$
⇒ 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 765 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 775 Marks
Prove the following statement by principle of mathematical induction:
$1 + 5 + 9 + ...... + (4n - 3) = n(2n - 1)$, for all natural numbers n.
AnswerLet 1 + 5 + 9 + ...... + (4n - 3) = n(2n - 1), $\forall\text{ n}\in\text{N}$
Step 1: P(1): L.H.S. = 4 × 1 - 3 = 1
R.H.S. = 1(2 × 1 - 1) = 1
L.H.S. = R.H.S. So P(1) is true.
Step 2: Assume P(k) is true for some $\text{k }\in\text{ N},\text{ k}\geq5\Rightarrow\text{P(k): }1+5+9+\ ......\ +(4\text{k}-3)=\text{k}(2\text{k}-1)\ ...(\text{i})$ is true.
Step 3: Now we have to prove $P(k + 1): 1 + 5 + ...... + (4k - 3) + [4(k + 1) - 3] = (k + 1)[2(k + 1) - 1]$ is true.
$\Rightarrow P(k + 1): 1 + 5 + 9 + ...... + (4k - 3) + (4k + 1) = (k + 1)(2k + 1)$ is true.
Adding (4k + 1) on both sides of equation $(i)$
$1 + 5 + 9 + ..... + (4k - 3) + (4k + 1)$
$= k(2k - 1) + (4k + 1) = 2k^2 - k + 4k + 1$
$= 2k^2+ 3k + 1 = 2k^2 + 2k + k + 1$
$= 2k(k + 1) + 1(k + 1) = (2k + 1)(k + 1)$
$\Rightarrow P(k + 1)$ is true.
Hence, $P(k + 1)$ is true whenever $P(k)$ is true.
Therefore by the principle of mathematical induction we have P(n) is true for all n.
View full question & answer→Question 785 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 795 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$S_n=1^2+2 \times 2^2+3^2+2 \times 4^2+\ldots$
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 $S _{ n }=\frac{ n ( n +1)^2}{2}$
$S_2=\frac{2 \times 9}{2}=9$
Now using induction we first shiw this is true also
$\text { for } n=3 \text {, we get } S_3=1+8+9=18$
From RHS, we have if n is odd $S _{ n }=\frac{ n ^2( n +1)}{2}$
$S_3=\frac{9 \times 4}{2}=18$
Lets assume above is true for $n=k$, we get
$K$ is even, $S_k=1^2+2 \times 2^2+3^2+2 \times 4^2+\ldots+2 \times k^2 \ldots$
$K$ is odd, $S_k=1^2+2 \times 2^2+3^2+2 \times 4^2+\ldots+k^2 \ldots$ (2)
Now lets prove for $n=k+1$
If $k$ is even, $k+1$ is odd we get
$S_{k+1}=1^2+2 \times 2^2+3^2+\ldots+2 \times k^2+(k+1)^2 \ldots$
From above relation, we get
$S_k=1^2+2 \times 2^2+3^2+2 \times 4^2+\ldots+2 \times k^2=\frac{k(k+1)^2}{2}$
Substitute this on 3 , we get
$S_{k+1}=\frac{k(k+1)^2}{2}+(k+1)^2=\frac{(k+1)^2(k+2)}{2}$
$=\text { RHS (when ' } k+1^{\prime} \text { ' is odd) }$
Hence Proved
View full question & answer→Question 805 Marks
Prove the following statement by principle of mathematical induction:
$n(n^2 + 5)$ is divisible by $6$, for each natural number n.
AnswerLet $P(n): n\left(n^2+5\right)$ is divisible by 6 , for each natural number.
Now $P(l): 1\left(l^2+5\right)=6$, which is divisible by 6 .
Hence, $P ( l )$ is true.
Let us assume that $P ( n )$ is true for some natural number
$n = k$. $P(k): k\left(k^2+5\right)$ is divisible by 6 .
Or $k\left(k^2+5\right)=6 m, m \in N \ldots$...(i)
Now, we have to prove that $P(k+1)$ is true.
$P(k + l): (k + l)[(k + l)^2 + 5]$
$= (k + l)[K^2 + 2K + 6]$
$= K^3 + 3K^2 + 8K + 6$
$= (K^2 + 5K) + 3K^2 + 3K + 6 = K(k^2 + 5) + 3(k^2 + k + 2)$
$= (6m) + 3(k^2 + k + 2) $(Using(i))
Now, $k^2+ k + 2$ is always even if $A$ is odd or even.
So, $3(K^2 + K + 2)$ is divisible by $6$ and hence, $(6m) + 3(k^2 + k + 2)$ is divisible by $6.$
Hence, $P(k + 1)$ is true whenever $P(k)$ is true.
So, by the principle of mathematical induction $P(n)$ is true for any natural number n.
View full question & answer→Question 815 Marks
Prove the following statement by principle of mathematical induction:
$n^3- n$ is divisible by 6, for each natural number $\text{n}\geq2$
AnswerLet $P(n): n^3 - n$
Step 1:$ P(2): 2^3 - 2 = 6$ which is divisible by 6. so, P(2) is true.
Step 2: Assume P(k) is true for some $\text{k }\in\text{ N}$ where $\text{k}\geq2\Rightarrow\text{P(k)}:\text{k}^3-\text{k}=6\lambda,\lambda\ \in\text{ N}$ is true.
${\text{P}(\text{k+1): }}(\text{k}+1)^3-(\text{k}+1)$
$=\text{k}^3+1+3\text{k}^2+3\text{k}-\text{k}-1$
$=6\lambda+\text{k}+1+3\text{k}^2+3\text{k}-\text{k}-1$ [from(i)]
$=6\lambda+3(\text{k}^2+\text{k})$
$=6\lambda+3\text{k}(\text{k}+1)$
We know that k(k + 1) = 2m since k(k + 1) is always even for every $\text{k }\in\text{ N}$
So, $\text{P}(\text{k}+1):6\lambda+3\cdot2\text{m}=6(\lambda+\text{m})$ which is divisible by 6.
$\Rightarrow\text{P}(\text{k}+1)$ is true.
Hence P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have P(n) is true for all $\text{n}\geq2$
View full question & answer→Question 825 Marks
Use the Principle of Mathematical Induction in the following Exercis.
Prove that: $\cos\theta \cdot\cos2\theta\cdot\cos2^2\theta\ ....\text{ cos}2^{\text{n}-1}\theta=\frac{\sin2^{\text{n}}\theta}{2^{\text{n}}\sin\theta},\forall\text{ n}\in\text{N}$
AnswerLet $\text{P(n)}:\cos\theta \cdot\cos2\theta\cdot\cos2^2\theta\ ....\text{ cos}2^{\text{n}-1}\theta=\frac{\sin2^{\text{n}}\theta}{2^{\text{n}}\sin\theta},\forall\text{ n}\in\text{N}$
Step 1: $\text{P(1)}:\cos\theta=\frac{\sin2^1\theta}{2^1\sin\theta}=\frac{\sin2\theta}{2\sin\theta}=\frac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$
P(1) is true.
Step 2: Assume P(k) is true for some $\text{k }\in\text{ N}$
$\text{P}(\text{k}):\cos\theta\cdot\cos2\theta\cdot\cos2^2\theta\ ....\text{cos }2^{\text{k}-1}\theta=\frac{\sin2^{\text{k}}\theta}{2^{\text{k}}\sin\theta}\ ......(\text{i})$ is true.
Step 3: Now we have to prove $\text{P}(\text{k}+1):\cos\theta\cdot\cos2\theta\cdot\cos2^2\theta\ ....\\\text{cos }2^{\text{k}-1}\theta\cdot\cos2^{(\text{k+1})-1}\theta=\frac{\sin2^{\text{k}+1}\theta}{2^{\text{k}+1}\sin\theta}$ is true.
Multiplying both sides of equation (i) by $\cos2^{\text{k}}\theta$
$\text{P}(\text{k}+1):\cos\theta\cdot\cos2\theta\cdot\cos2^2\theta\ ....\\\text{cos }2^{\text{k}-1}\theta\cdot\cos2^{\text{k}}\theta=\frac{\sin2^{\text{k}}\theta}{2^{\text{k}}\sin\theta}\cdot\cos2^{\text{k}}\theta$
$=\frac{2\sin2^{\text{k}}\theta\cos2^{\text{k}}\theta}{2\cdot2^{\text{k}}\sin\theta}$
$=\frac{\sin2\cdot2^{\text{k}}\theta}{2^{\text{k}+1}\sin\theta}$ $\big[\because2\sin\theta\cos\theta=\sin2\theta\big]$
$=\frac{\sin2^{\text{k}+1}\theta}{2^{\text{k}+1}\sin\theta}$
$\text{P}(\text{k}+1)$ is true.
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have P(n) is true for all n.
View full question & answer→Question 835 Marks
Prove the following statement by principle of mathematical induction:2n < (n + 2)! for all natural number n.
AnswerLet P(n): 2n < (n + 2)! for all natural numbers n.
P(l): 2 < (1 + 2)! or 2 < 3! or 2 < 6, which is true.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 2k < (k + 2)! .....(i)
To prove that P(k + 1) is true, we have to show that
P(k + 1): 2(k + l) < (k + 1 + 2)!
Or 2(k + 1) < (k + 3)!
Using (i) we have,
2(k + 1) = 2k + 2 < (k + 2)! + 2 ....(ii)
Now let, (k + 2)! + 2 < (k + 3)! .....(iii)
⇒ 2 < (k + 3)! - (k + 2)!
⇒ 2 < (k + 2) ! [k + 3 - 1]
⇒ 2 < (k + 2) ! (k + 2), which is true for any natural number.
Using (ii) and (iii) we get 2(k + 1) < (k + 3)!
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.
View full question & answer→Question 845 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\text{n}-1)(3\text{n}+2)}=\frac{\text{n}}{(6\text{n}+4)}.$
AnswerLet the given statement be P(n), i.e., $\text{P(n)}:\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\text{n}-1)(3\text{n}+2)}=\frac{\text{n}}{(6\text{n}+4)}$ For n = 1, we have$\text{P(1)}=\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}=\frac{1}{10},$
which is true. Let P(k) be true for some positive integer k, i.e., $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\text{k}-1)(3\text{k}+2)}=\frac{\text{k}}{(6\text{k}+4)}\ \ ...(\text{i})$ We shall now prove that P(k + 1) is true. Consider, $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\text{k}-1)(3\text{k}+2)}+\frac{1}{\{3(\text{k}+1)-1\}\{3(\text{k+1})+2\}}$ $=\frac{\text{k}}{(6\text{k}+4)}+\frac{1}{(3\text{k}+3-1)(3\text{k+3+2})}\ \ [\text{Using (i)}]$ $=\frac{\text{k}}{2(3\text{k}+2)}+\frac{1}{(3\text{k}+2)(3\text{k+5})}$ $=\frac{1}{(3\text{k}+2)}\Big(\frac{\text{k}}{2}+\frac{1}{3\text{k}+5}\Big)$ $=\frac{1}{(3\text{k}+2)}\Big(\frac{3\text{k}^2+5\text{k}+2}{2(3\text{k}+5)}\Big)$ $=\frac{1}{(3\text{k}+2)}\Big(\frac{(3\text{k}+2)(\text{k}+1)}{2(3\text{k}+5)}\Big)$ $=\frac{(\text{k+1})}{6\text{k}+10}$ $=\frac{(\text{k+1})}{6(\text{k+1})+4}$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 855 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1.2.3+2.3.4+...+\text{n(n+1)(n+2)}=\frac{\text{n(n+1)(n+2)(n+3)}}{4}.$
AnswerLet the given satement be P(n), i.e., $\text{P(n)}:1.2.3+2.3.4+...+\text{n(n+1)(n+2)}=\frac{\text{n(n+1)(n+2)(n+3)}}{4}$ For n = 1, we have $\text{P}(1):1.2.3=6=\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1.2.3.4}{4}=6$ which is true. Let P(k) be true for some positive integer k, i.e., $1.2.3+2.3.4+...+\text{k(k+1)(k+2)}=\frac{\text{k(k+1)(k+2)(k+3)}}{4}\ \ ...(\text{i})$ We shall now prove that P(k + 1) is true. Consider, 1.2.3 + 2.3.4 + .... + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3) = {1.2.3 + 2.3.4 + ... + k(k + 1)(k + 2)} + (k + 1)(k + 2)(k + 3)$=\frac{\text{k(k+1)(k+2)(k+3)}}{4}+(\text{k+1)(k+2)(k+3)}\ \ [\text{Using (i)}]$
$=(\text{k+1)(k+2)(k+3)}\Big(\frac{\text{k}}{4}+1\Big)$
$=\ \frac{(\text{k+1)(k+2)(k+3)(k+4)}}{4}$
$=\ \frac{(\text{k+1)(k+1+1)(k+1+2)(k+1+3)}}{4}$
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 865 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:
$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2\text{n}+1)(2\text{n}+3)}=\frac{\text{n}}{3(2\text{n}+3)}.$
AnswerLet the given statement be P(n), i.e., $\text{P(n)}:\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2\text{n}+1)(2\text{n}+3)}=\frac{\text{n}}{3(2\text{n}+3)}$ For n = 1, we have $\text{P(1)}:\frac{1}{3.5}=\frac{1}{3(2.1+3)}=\frac{1}{3.5},$ which is true. Let P(k) be true for some positive integer k, i.e., $\text{P(k)}:\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2\text{k}+1)(2\text{k}+3)}=\frac{\text{k}}{3(2\text{k}+3)}\ ....(1)$ We shall now prove that P(k + 1) is true. Consider$\Big[\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2\text{k}+1)(2\text{k}+3)}\Big]+\frac{1}{\{2(\text{k+1})+1\}\{2(\text{k}+1)+3\}}$
$=\frac{\text{k}}{3(2\text{k}+3)}+\frac{1}{(2\text{k}+3)(2\text{k}+5)}\ \ [\text{Using (1)}]$
$=\frac{1}{(2\text{k}+3)}\Big[\frac{\text{k}}{3}+\frac{1}{(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}\Big[\frac{\text{k}(2\text{k+5})+3}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}\Big[\frac{2\text{k}^2+5\text{k}+3}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k}+3)}\Big[\frac{2\text{k}^2+2\text{k}+3\text{k}+3}{3(2\text{k}+5)}\Big]$
$=\frac{1}{(2\text{k+3})}\Big[\frac{2\text{k}(\text{k+1})+3(\text{k+1})}{3(2\text{k+5})}\Big]$
$=\frac{(\text{k+1})(2\text{k+3})}{3(2\text{k}+3)(2\text{k+5})}$
$=\frac{(\text{k+1})}{3\{2(\text{k+1})+3\}}$
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 875 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:
$10^{2\text{n}–1} + 1$ is divisible by 11.
Answer$\text{P(n)}:10^{2\text{n-1}}+1$ is divisible by 11.It can be observed that P(n) is true for n = 1 since
$\text{P(1)}=10^{2.1-1}+1=11,$ which is divisible by 11.
Let P(k) be true for some positive integer k, i.e.,
$10^{2\text{k}-1}+1$ is divisible by 11.
$\therefore\ 10^{2\text{k}-1}+1=11\text{m, where m}\in\text{N}\ ...(1)$
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider,
$=10^{2(\text{k}+1)-1}+1$
$=10^{2\text{k}+2-1}+1$
$=10^{2\text{k}+1}+1$
$=10^2(10^{2\text{k}-1}+1-1)+1$
$=10^2(10^{2\text{k}-1}+1)-10^2+1$
$=10^2.11\text{m}-100+1\ \ [\text{Using (1)}]$
$=100\times11\text{m}-99$
$=11(100\text{m}-9)$
$=11\text{r, where r =}(100\text{m}-9)$ is some natural number
Therefore, $10^{2(\text{k}+1)-1}+1$ is divisible by 11.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 885 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:
$3^{2\text{n}+2}-8\text{n}-9$ is divisible by 8.
AnswerLet the given statement be P(n), i.e.,$\text{P(n)}:3^{2\text{n}+2}-8\text{n}-9$ is divisible by 8.
It can be observed that P(n) is true for n = 1 since $3^{2\times1+2}-8\times1-9=64,$ which is divisible by 8.Let P(k) be true for some positive integer k, i.e.,
$3^{2\text{k}+2}-8\text{k}-9$ is divisible by 8.
$\therefore\ 3^{2\text{k}+2}-8\text{k}-9=8\text{m};\ \text{where m}\in\text{N}\ ....(1)$
We shall now prove that P(k + 1) is true whenever P(k) is true. Consider, $3^{2(\text{k+1})+2}-8(\text{k+1})-9$ $=3^{2\text{k+2}}.3^2-8\text{k}-8-9$ $=3^2(3^{2\text{k+2}}-8\text{k}-9+8\text{k}+9)-8\text{k}-17$ $=3^2(3^{2\text{k+2}}-8\text{k}-9)+3^2(8\text{k}+9)-8\text{k}-17$ $=9.8\text{m}+9(8\text{k}+9)-8\text{k}-17$ $=9.8\text{m}+72\text{k}+81-8\text{k}-17$ $=9.8\text{m}+64\text{k}+64$ $=8(9\text{m}+8\text{k}+8)$ Therefore, $3^{2(\text{k+1})+2}-8(\text{k+1})-9$ is divisible by 8. Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 895 Marks
Use the Principle of Mathematical Induction in the following Exercis.
Prove that for all $\text{n}\in\text{N},\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ ....\ +\cos\big[\alpha+(\text{n}-1)\beta\big]$$=\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}}$
AnswerLet $\text{P(n):}\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ ....\ +\cos\big[\alpha+(\text{n}-1)\beta\big]$
$=\frac{\cos\Big[\alpha+\big(\frac{\text{n}-1}{2}\big)\beta\Big]\sin\big(\frac{\text{n}\beta}{2}\big)}{\sin\frac{\beta}{2}}$
Now, ${\text{P}(1){:}}\cos\alpha=\frac{\cos\Big[\alpha+\big(\frac{\text{1}-1}{2}\big)\beta\Big]\sin\frac{\beta}{2}}{\sin\frac{\beta}{2}}=\frac{\cos\alpha\ {\sin\frac{\beta}{2}}}{{\sin\frac{\beta}{2}}}=\cos\alpha$
Hence, P(1) is true.
Now, let us assume that P(n) is true for some natural number n = k.
$\therefore\ \text{P(k):}\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ ....\ +\cos\big[\alpha+(\text{k}-1)\beta\big]$
$=\frac{\cos\Big[\alpha+\big(\frac{\text{k}-1}{2}\big)\beta\Big]\sin\big(\frac{\text{k}\beta}{2}\big)}{\sin\frac{\beta}{2}}\ ....(\text{i})$
Now, to prove that P(k + 1) is true, we have to show that,
${\text{P}(\text{k}+1){: }}\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ ....\\\ +\cos\big[\alpha+(\text{k}-1)\beta\big]+\cos\big[\alpha+(\text{k}+1-1)\beta\big]$
$=\frac{\cos\big(\alpha+\frac{\text{k}\beta}{2}\big)\sin\frac{(\text{k}+1)\beta}{2}}{\sin\frac{\beta}{2}}$
$\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ ......\\\ +\cos\big[\alpha+(\text{k}-1)\beta\big]+\cos(\alpha+\text{k}\beta)$
$=\frac{\cos\Big[\alpha+\big(\frac{\text{k}-1}{2}\big)\Big]\sin\frac{\text{k}\beta}{2}}{\sin\frac{\beta}{2}}+\cos(\alpha+\text{k}\beta)$ [Using(i)]
$=\frac{\cos\Big[\alpha+\big(\frac{\text{k}-1}{2}\big)\Big]\sin\frac{\text{k}\beta}{2}+\cos(\alpha+\text{k}\beta)\sin\frac{\beta}{2}}{\sin\frac{\beta}{2}}$
$=\frac{\sin\Big(\alpha+\frac{\text{k}\beta}{2}-\frac{\beta}{2}+\frac{\text{k}\beta}{2}\Big)-\sin\Big(\alpha+\frac{\text{k}\beta}{2}-\frac{\beta}{2}-\frac{\text{k}\beta}{2}\Big)\\+\sin\Big(\alpha+\text{k}\beta+\frac{\beta}{2}\Big)-\sin\Big(\alpha+\text{k}\beta-\frac{\beta}{2}\Big)}{2\sin\frac{\beta}{2}}$
$=\frac{\sin\Big(\alpha+\text{k}\beta+\frac{\beta}{2}\Big)-\sin\Big(\alpha-\frac{\beta}{2}\Big)}{2\sin\frac{\beta}{2}}$
$=\frac{2\cos\frac{1}{2}\Big(\alpha+\text{k}\beta+\frac{\beta}{2}+\alpha-\frac{\beta}{2}\Big)\sin\frac{1}{2}\Big(\alpha+\text{k}\beta+\frac{\beta}{2}-\alpha+\frac{\beta}{2}\Big)}{2\sin\frac{\beta}{2}}$
$=\frac{\cos\Big(\frac{2\alpha+\text{k}\beta}{2}\Big)\sin\Big(\frac{\text{k}\beta+\beta}{2}\Big)}{\sin\frac{\beta}{2}}=\frac{\cos\Big(\alpha+\frac{\text{k}\beta}{2}\Big)\sin(\text{k}+1)\frac{\beta}{2}}{\sin\frac{\beta}{2}}$
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.
View full question & answer→Question 905 Marks
Prove the following by using the principle of mathematical induction for all $n \in N:$
$\text{x}^{2\text{n}}-\text{y}^{2\text{n}}$ is divisible by $x + y.$
AnswerLet the given statement be $P(n)$, i.e., $P(n)$ : $x^{2 n}-y^{2 n}$ is divisible by $x+y$.
It can be observed that $P(n)$ is true for $n=1$.
This is so because $x^{2 \times 1}-y^{2 \times 1}=x^2-y^2=(x+y)(x-y)$ is divisible by $(x+y)$.
Let $P(k)$ be true for some positive integer $k$, i.e., $x^{2 k}-y^{2 k}$ is divisible by $x+y$.
$\therefore x^{2 k}-y^{2 k}=m(x+y) \text {, where } m \in N \ldots$
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Consider,
$\text{x}^{2(\text{k+1})}-\text{y}^{2(\text{k}+1)}$
$=\text{x}^{2\text{k}}.\text{x}^2-\text{y}^{2\text{k}}.\text{y}^2$
$=\text{x}^2(\text{x}^{2\text{k}}-\text{y}^{2\text{k}}+\text{y}^{2\text{k}})-\text{y}^{2\text{k}}.\text{y}^2$
$=\text{x}^2\{\text{m}(\text{x+y)}+\text{y}^{2\text{k}}\}-\text{y}^{2\text{k}}.\text{y}^2\ \ [\text{Using (1)}]$
$=\text{m(x+y)x}^2+\text{y}^{2\text{k}}.\text{x}^2-\text{y}^{2\text{k}}.\text{y}^2$
$=\text{m(x+y)x}^2+\text{y}^{2\text{k}}(\text{x}^2-\text{y}^2)$
$=\text{m(x+y)x}^2+\text{y}^{2\text{k}}(\text{x}+\text{y})(\text{x}-\text{y})$
$=(\text{x+y})\{\text{mx}^2+\text{y}^{2\text{k}}(\text{x}-\text{y})\},$ which is a factor of (x + y).
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 915 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1^3+2^3+3^3+....+\text{n}^3=\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2.$
AnswerLet the given statement be P(n), i.e.,$\text{P(n)}:1^3+2^3+3^3+....+\text{n}^3=\Big(\frac{\text{n}(\text{n+1)}}{2}\Big)^2$
For n = 1, we have $\text{P}(1):1^3=1=\Big(\frac{1(1+1)}{2}\Big)^2=\Big(\frac{1.2}{2}\Big)^2=1^2=1$ which is true. Let P(k) be true for some positive integer k, i.e.,$1^3+2^3+3^3+....+\text{k}^3=\Big(\frac{\text{k(k+1)}}{2}\Big)^2\ \ ....(\text{i})$
We shall now prove that P(k + 1) is true. Consider, $1^3+2^3+3^3+...+\text{k}^3+(\text{k}+1)^3$ $=(1^3+2^3+3^3+...+\text{k}^3)+(\text{k}+1)^3$ $=\Big(\frac{\text{k(k+1)}}{2}\Big)^2+(\text{k}+1)^3\ \ [\text{Using (i)}]$ $=\frac{\text{k}^2(\text{k+1})^2}{4}+(\text{k}+1)^3$ $=\frac{\text{k}^2(\text{k+1})^2+4(\text{k}+1)^3}{4}$ $=\frac{(\text{k+1})^2\{\text{k}^2+4(\text{k+1})\}}{4}$ $=\frac{(\text{k+1})^2\{\text{k}^2+4\text{k+4}\}}{4}$ $=\frac{(\text{k}+1)^2(\text{k}+2)^2}{4}$ $=\frac{(\text{k}+1)^2(\text{k}+1+1)^2}{4}$ $=\Big(\frac{(\text{k+1})(\text{k+1+1})}{2}\Big)^2$ Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural number i.e., n.
View full question & answer→Question 925 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1.2+2.3+3.4+...+\text{n.(n+1)}=\Big[\frac{\text{n(n+1)(n+2)}}{3}\Big].$
AnswerLet the given statement be P(n), i.e.,
$\text{P(n)}:1.2+2.3+3.4+...+\text{n.(n+1)}=\Big[\frac{\text{n(n+1)(n+2)}}{3}\Big]$
For n = 1, we have
$\text{P(1)}:1.2=2=\frac{1(1+1)(1+2)}{3}=\frac{1.2.3}{3}=2,$
which is true.
Let P(k) be true for some positive integer k, i.e.,
$1.2+2.3+3.4+...+\text{k}.(\text{k+1})=\Big[\frac{\text{k(k+1)(k+2)}}{3}\Big]\ \ ....(\text{i})$
We shall now prove that P(k + 1) is true.
Consider,
$1.2+2.3+3.4+....+\text{k.(k+1)+(k+1).(k+2)}$
$=[1.2+2.3+3.4+....+\text{k.(k+1)]+(k+1).(k+2)}$
$=\frac{\text{k(k+1)(k+2)}}{3}+(\text{k+1)(k+2)}\ \ [\text{Using (i)}]$
$=(\text{k+1)(k+2)}\Big(\frac{\text{k}}{3}+1\Big)$
$=\frac{(\text{k+1)(k+2)(k+3)}}{3}$
$=\frac{(\text{k+1)(k+1+1)(k+1+2)}}{3}$
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 935 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1.2+2.2^2+3.2^3+...+\text{n}.2^\text{n}=(\text{n}-1)2^{\text{n+1}}+2.$
AnswerLet the given statement be P(n), i.e.,
$\text{P(n)}:1.2+2.2^2+3.2^3+...+\text{n}.2^\text{n}=(\text{n}-1)2^{\text{n+1}}+2$
For n = 1, we have
$\text{P(1)}:1.2=2=(1-1)2^{1+1}+2=0+2=2,$
which is true.
Let P(k) be true for some positive integer k, i.e.,
$1.2+2.2^2+3.2^3+...+\text{k}.2^\text{k}=(\text{k}-1)2^{\text{k}+1}+2\ \ ....(\text{i})$
We shall now prove that P(k + 1) is true.
Consider,
$\{1.2+2.2^2+3.2^3+...+\text{k}.2^{\text{k}}\}+(\text{k}+1).2^{\text{k}+1}$
$=(\text{k}-1)2^{\text{k}+1}+2+(\text{k}+1)2^{\text{k+1}}$
$=2^{\text{k+1}}\{(\text{k}-1)+(\text{k}+1)\}+2$
$=2^{\text{k}+1}.2\text{k}+2$
$=\text{k}.2^{(\text{k}+1)+1}+2$
$=\{(\text{k+1})-1\}2^{(\text{k}+1)+1}+2$
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
View full question & answer→Question 945 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1.3+3.5+5.7+...+(2\text{n}-1)(2\text{n}+1)=\frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}.$
Answer$\text{Let P(n)}=1.3+3.5+5.7+...+(2\text{n}-1)(2\text{n}+1)=\frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$For n = 1
$\text{P}(1)=(2\times1-1)(2\times1+1)=\frac{1[4(1)^2+6\times1-1]}{3}$
⇒ 3 = 3
$\therefore$ P(1) is true.
Now, let P(n) be true for n = k
$\therefore\ \text{P(k)}=1.3+3.5+5.7+...+(2\text{k}-1)(2\text{k}+1)=\frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3}$
For n = k + 1
$\text{P(k+1)}=1.3+3.5+5.7+...+(2\text{k}-1)(2\text{k}+1)+[2(\text{k}+1)-1][2(\text{k}+1)+1]$
$=\frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3}+[2(\text{k+1})-1][2(\text{k+1})+1]$
$\Rightarrow\ \text{P(k+1)}=\frac{4\text{k}^3+6\text{k}^2-\text{k}}{3}+(2\text{k}+1)(2\text{k}+3)$
$=\frac{4\text{k}^3+6\text{k}^2-\text{k}+3(4\text{k}^2+8\text{k}+3)}{3}$
$\Rightarrow\ \text{P(k+1)}=\frac{4\text{k}^3+6\text{k}^2-\text{k}+12\text{k}^2+24\text{k}+9}{3}$
$=\frac{4\text{k}^3+18\text{k}^2+23\text{k}+9}{3}$
$=\frac{(\text{k+1})(4\text{k}^2+14\text{k}+9)}{3}$
$\therefore$ P(k + 1) is true.
Therefore, P(k) is true.
⇒ P(k + 1) is true.
Hence by Principle of Mathematical Induction, P(n) is true for all $\text{n}\in\text{N.}$
View full question & answer→Question 955 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...\text{n})}=\frac{2\text{n}}{(\text{n+1})}.$
Answer$\text{Let}\ \text{P(n)}=1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+\text{n})}=\frac{2\text{n}}{(\text{n+1})}$For n = 1,
$\text{P(1)}=1=\frac{2\times1}{1+1}$
⇒ 1 = 1
$\therefore$ P(1) is true.
Now, let P(n) be true for n = k
$\therefore\ \text{P(k)}=1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+\text{k})}=\frac{2\text{k}}{(\text{k+1})}\ ....(\text{i})$
For n = k + 1,
$\text{P(k+1)}=1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+\text{k})}$
$+\frac{1}{(1+2+3+...+\text{k+1})}=\frac{2\text{k}}{(\text{k+1})}+\frac{1}{(1+2+3+...+\text{k+1})}\ \ [\text{Using (i)}]$
$\Rightarrow\ \text{P(k+1)}=\frac{2\text{k}}{\text{k}+1}+\frac{1}{\frac{(\text{k+1})(\text{k+2})}{2}}$
$=\frac{2\text{k}}{\text{k+1}}+\frac{2}{(\text{k+1})(\text{k+2})}$
$=\frac{2}{\text{k+1}}\Big[\frac{(\text{k+1})^2}{\text{k}+2}\Big]=\frac{2(\text{k+1})}{\text{k+2}}$
$\therefore$ P(k + 1) is true.
Therefore, P(k) is true.
⇒ P(k + 1) is true.
Hence by Principle of Mathematical Induction, P(n) is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 965 Marks
Prove the following by using the principle of mathematical induction for all n ∈ N:$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\text{n(n+1)(n+2)}}=\frac{\text{n(n+3)}}{4(\text{n+1})(\text{n}+2)}.$
Answer$\text{Let }\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\text{n(n+1)(n+2)}}=\frac{\text{n(n+3)}}{4(\text{n+1})(\text{n}+2)}$For n = 1
$\text{P(1)}=\frac{1}{1(1+1)(1+2)}=\frac{1(1+3)}{4(1+1)(1+2)}$
$\Rightarrow\ \frac{1}{6}=\frac{1}{6}$
$\therefore$ P(1) is true.
Now, let P(n) be true for n = k
$\therefore\ \text{P(k)}=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\text{k(k+1)(k+2)}}=\frac{\text{k(k+3)}}{4(\text{k+1})(\text{k}+2)}\ \ ...(\text{i})$
For n = k + 1
$\Rightarrow\ \text{R.H.S.}=\frac{(\text{k+1})(\text{k+4})}{4(\text{k+2})(\text{k+3})}$
And $\text{L.H.S.}=\frac{\text{k(k+3)}}{4(\text{k+2})(\text{k}+3)}+\frac{1}{(\text{k+1})(\text{k+2})(\text{k+3})}\ \ [\text{Using (i)}]$
$=\frac{1}{(\text{k+1})(\text{k+2})}\Big[\frac{\text{k}^2+3\text{k}}{4}+\frac{1}{\text{k}+3}\Big]$
$=\frac{1}{(\text{k+1})(\text{k+2})}\Big[\frac{\text{k}^3+6\text{k}^2+9\text{k}+4}{4(\text{k+3})}\Big]$
$=\frac{1}{(\text{k+1})(\text{k+2})}\Big[\frac{(\text{k+1)}^2(\text{k+4)}}{4(\text{k+3})}\Big]$
$=\frac{(\text{k+1})(\text{k+4})}{4(\text{k+2})(\text{k}+3)}$
$\therefore$ P(k + 1) is true.
Therefore, P(k) is true.
⇒ P(k + 1) is true.
Hence by Principle of Mathematical Induction, P(n) is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 975 Marks
Use the Principle of Mathematical Induction in the following Exercis.
Prove that: $\sin\theta+\sin2\theta+\sin3\theta+\ ....\ +\sin\text{n}\theta=\frac{\sin\frac{\text{n}\theta}{2}\cdot\sin\frac{\text{n}+1}{2}\theta}{\sin\frac{\theta}{2}},$ for all $\text{n}\in\text{N}.$
AnswerLet $\text{P(n)}:\sin\theta+\sin2\theta+\sin3\theta+\ ....\ +\sin\text{n}\theta$
$=\frac{\sin\frac{\text{n}\theta}{2}\cdot\sin\frac{\text{n}+1}{2}\theta}{\sin\frac{\theta}{2}}$
Step 1: $\text{P}(1):\text{L.H.S.}=\sin\theta\ \ \text{R.H.S.}=\frac{\sin\frac{\theta}{2}\cdot\sin\theta}{\sin\frac{\theta}{2}}=\sin\theta$
$\text{L.H.S.}=\text{R.H.S.}\ \ \ \therefore\ \text{P}(1)$ is true.
Step 2: Assume P(k) is true for some $\text{k }\in\text{ N}$
$\text{P}(\text{k}):\sin\theta+\sin2\theta+\sin3\theta+\ ....\ +\sin\text{k}\theta$
$=\frac{\sin\frac{\text{k}\theta}{2}\cdot\sin\big(\frac{\text{k}+1}{2}\big)\theta}{\sin\frac{\theta}{2}}\ ....(\text{i})$ is true.
Step 3: Now we have to prove
$\text{P}(\text{k}+1):\sin\theta+\sin2\theta+\sin3\theta+\ .....\ +\sin(\text{k}+1)\theta$
$=\frac{\sin\frac{\text{k}\theta}{2}\cdot\sin\big(\frac{\text{k}+1}{2}\big)\theta}{\sin\frac{\theta}{2}}+\sin(\text{k}+1)\theta$
$=\frac{\sin\frac{\text{k}\theta}{2}\cdot\sin\big(\frac{\text{k}+1}{2}\big)\theta+\sin(\text{k}+1)\theta\cdot\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}}$
$=\frac{\sin\big(\frac{\text{k}\theta}{2}\big)\cdot\sin\big(\frac{\text{k}+1}{2}\big)\theta+2\sin(\text{k}+1)\theta\cdot\sin\big(\frac{\theta}{2}\big)}{2\sin\big(\frac{\theta}{2}\big)}$
$\big[\because\ 2\sin\text{A}\sin\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A}+\text{B})\big]$
$=\frac{\cos\big[\frac{\text{k}\theta}{2}-\big(\frac{\text{k}+1}{2}\big)\theta\big]-\cos\big[\frac{\text{k}\theta}{2}-\big(\frac{\text{k}+1}{2}\big)\theta\big]+\cos\big[(\text{k}+1)\theta-\frac{\theta}{2}\big]-\cos\big[(\text{k}+1)\theta+\frac{\theta}{2}\big]}{2\sin\big(\frac{\theta}{2}\big)}$
$=\frac{\cos\big(\frac{-\theta}{2}\big)-\cos\big(\text{k}\theta+\frac{\theta}{2}\big)+\cos\big(\text{k}\theta+\frac{\theta}{2}\big)-\cos\big(\text{k}\theta+\frac{3\theta}{2}\big)}{2\sin\big(\frac{\theta}{2}\big)}$
$=\frac{\cos\big(\frac{\theta}{2}\big)-\cos\big(\text{k}\theta+\frac{3\theta}{2}\big)}{2\sin\frac{\theta}{2}}$
$=\frac{-2\sin\Bigg(\frac{\frac{\theta}{2}+\text{k}\theta+\frac{3\theta}{2}}{2}\Bigg)\cdot\sin\Bigg(\frac{\frac{\theta}{2}-\text{k}\theta-\frac{3\theta}{2}}{2}\Bigg)}{2\sin\frac{\theta}{2}}$
$\Big[\because\ \cos\text{A}-\cos\text{B}=-2\sin\frac{(\text{A}+\text{B})}{2}\sin\frac{(\text{A}-\text{B})}{2}\Big]$
$=\frac{-2\sin\big(\frac{\text{k}\theta+2\theta}{2}\big)\cdot\sin\big(\frac{-\text{k}\theta-\theta}{2}\big)}{2\sin\frac{\theta}{2}}$ $\big[\because\ \sin(-\theta)=-\sin\theta\big]$
$=\frac{\sin\big[\frac{(\text{k}+1)+1}{2}\big]\cdot\sin\big[\frac{\text{k}+1}{2}\big]\theta}{\sin\frac{\theta}{2}}$
P(k + 1) is true.
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principl of mathematical induction we have P(n) is true for all n.
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