NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

Question:2 Prove the following by using the principle of mathematical induction for all $n\in N$ : $1^3+2^3+3^3+...+n^3={\left(\frac{n(n+1)}{2}\right)}^2$

Answer:

Let the given statement be p(n) i.e.
$p(n):1^3+2^3+3^3+...+n^3={\left(\frac{n(n+1)}{2}\right)}^2$
For n = 1 we have
$p(1):1={\left(\frac{1(1+1)}{2}\right)}^2={\left(\frac{1(2)}{2}\right)}^2=(1{)}^2=1$ , which is true

For n = k we have
$p(k):1^3+2^3+3^3+...+k^3={\left(\frac{k(k+1)}{2}\right)}^2-(i)$ , Let's assume that this statement is true

Now,
For n = k + 1 we have
$p(k+1):1^3+2^3+3^3+...+(k+1{)}^3=1^3+2^3+3^3+...+k^3+(k+1{)}^3$
$=(1^3+2^3+3^3+...+k^3)+(k+1{)}^3$
$={\left(\frac{k(k+1)}{2}\right)}^2+(k+1{)}^3(using(i))$
$=\frac{k^2(k+1{)}^2+4(k+1{)}^3}{4}$
$=\frac{(k+1{)}^2(k^2+4(k+1))}{4}$
$=\frac{(k+1{)}^2(k^2+4k+4)}{4}$
$=\frac{(k+1{)}^2(k+2{)}^2}{4}(\because a^2+2ab+b^2=(a+b{)}^2)$
$={\left(\frac{(k+1)(k+2)}{2}\right)}^2$
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 n

Question:3 Prove the following by using the principle of mathematical induction for all $n\in N$ :

$1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+n)}=\frac{2n}{(n+1)}$

Answer:

Let the given statement be p(n) i.e.
$p(n):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+n)}=\frac{2n}{(n+1)}$
For n = 1 we have
$p(1):1=\left(\frac{2(1)}{1+1}\right)=\left(\frac{2}{2}\right)=1$ , which is true

For n = k we have
$p(k):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)}=\frac{2k}{(k+1)}-(i)$ , Let's assume that this statement is true

Now,
For n = k + 1 we have
$p(k+1):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k+1)}$ $=(1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)})+\frac{1}{(1+2+3+...+k+k+1)}$

$=\frac{2k}{k+1}+\frac{1}{(1+2+3+...+k+(k+1))}(using(i))$
$=\frac{2k}{k+1}+\frac{1}{\frac{(k+1)(k+1+1)}{2}}(\because 1+2+....+n=\frac{n(n+1)}{2})$
$=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}$
$=\frac{2}{k+1}(k+\frac{1}{k+2})$
$=\frac{2}{k+1}\left(\frac{k^2+2k+1}{k+2}\right)$
$=\frac{2}{k+1}.\frac{(k+1{)}^2}{k+2}$
$=\frac{2(k+1)}{k+2}$
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 n

Question:4 Prove the following by using the principle of mathematical induction for all $n\in N$ : $1.2.3+2.3.4+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

Answer:

Let the given statement be p(n) i.e.
$p(n):1.2.3+2.3.4+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
For n = 1 we have
$p(1):6=\left(\frac{1(1+1)(1+2)(1+3)}{4}\right)=\left(\frac{1.2.3.4}{4}\right)=6$ , which is true

For n = k we have
$p(k):1.2.3+2.3.4+...+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4}-(i)$ , Let's assume that this statement is true

Now,
For n = k + 1 we have
$p(k+1):1.2.3+2.3.4+...+k(k+1)(k+2)+(k+1)(k+2)(k+3)$ $=(1.2.3+2.3.4+...+k(k+1)(k+2))+(k+1)(k+2)(k+3)$

$=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)(using(i))$
$=\frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)}{4}$
$=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$
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 n

Question:5 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $1.3+{2.3}^2+{3.3}^3+...+n{.3}^n=\frac{(2n-1)3^{n+1}+3}{4}$

Answer:

Let the given statement be p(n) i.e.
$p(n):1.3+{2.3}^2+{3.3}^3+...+n{.3}^n=\frac{(2n-1)3^{n+1}+3}{4}$
For n = 1 we have
$p(1):3=\frac{(2(1)-1)3^{1+1}+3}{4}=\frac{(2-1)9+3}{4}=\frac{12}{4}=3$ , which is true

For n = k we have

$p(k):1.3+{2.3}^2+{3.3}^3+...+k{.3}^k=\frac{(2k-1)3^{k+1}+3}{4}-(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1.3+{2.3}^2+{3.3}^3+...+(k+1){.3}^{(k+1)}$$=1.3+{2.3}^2+{3.3}^3+...+k{.3}^k+(k+1){.3}^{(k+1)}$

$=\frac{(2k-1)3^{k+1}+3}{4}+(k+1){.3}^{(k+1)}$
$=\frac{(2k-1)3^{k+1}+3+4(k+1){.3}^{(k+1)}}{4}$
$=\frac{3^{k+1}((2k-1)+4(k+1))+3}{4}$
$=\frac{3^{k+1}(6k+3)+3}{4}$
$=\frac{3^{k+1}.3(2k+1)+3}{4}$
$=\frac{(2k+1)3^{k+2}+3}{4}$
$=\frac{(2(k+1)-1)3^{(k+1)+1}+3}{4}$
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 n

Question:6 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $1.2+2.3+3.4+...+n.(n+1)=\left[\frac{n(n+1)(n+2)}{3}\right]$

Answer:

Let the given statement be p(n) i.e.
$p(n):1.2+2.3+3.4+...+n.(n+1)=\left[\frac{n(n+1)(n+2)}{3}\right]$
For n = 1 we have
$p(1):2=\left[\frac{1(1+1)(1+2)}{3}\right]=\frac{1.2.3}{3}=2$ , which is true

For n = k we have

$p(k):1.2+2.3+3.4+...+k.(k+1)=\left[\frac{k(k+1)(k+2)}{3}\right]-(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1.2+2.3+3.4+...+(k+1).(k+2)$$=1.2+2.3+3.4+...+k(k+1)+(k+1).(k+2)$

$=\frac{k(k+1)(k+2)}{3}+(k+1).(k+2)(using(i))$
$=\frac{k(k+1)(k+2)+3(k+1).(k+2)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)(k+1+1)(k+1+2)}{3}$

Thus, p(k+1) is true whenever p(k) is true
Hence, by principle of mathematical induction , statement p(n) is true for all natural numbers n

Question:7 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$

Answer:

Let the given statement be p(n) i.e.
$p(n):1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$
For n = 1 we have
$p(1):1.3=3=\frac{1(4(1{)}^2+6(1)-1)}{3}=\frac{4+6-1}{3}=\frac{9}{3}=3$ , which is true

For n = k we have

$p(k):1.3+3.5+5.7+...+(2k-1)(2k+1)=\frac{k(4k^2+6k-1)}{3}-(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1.3+3.5+5.7+...+(2(k+1)-1)(2(k+1)+1)$$=1.3+3.5+5.7+...+(2k-1)(2k+1)+(2(k+1)-1)(2(k+1)+1)$

$=\frac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3)(using(i))$
$=\frac{k(4k^2+6k-1)+3(2k+1)(2k+3)}{3}$
$=\frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}$
$=\frac{(4k^3+6k^2-k+12k^2+28k+9)}{3}$
$=\frac{(4k^3+18k^2+23k+9)}{3}$
$=\frac{(4k^3+14k^2+9k+4k^2+14k+9)}{3}$
$=\frac{(k(4k^2+14k+9)+4k^2+14k+9)}{3}$
$=\frac{(4k^2+14k+9)(k+1)}{3}$
$=\frac{(k+1)(4k^2+8k+4+6k+6-1)}{3}$
$=\frac{(k+1)(4(k^2+2k+1)+6(k+1)-1)}{3}$
$=\frac{(k+1)(4(k+1{)}^2+6(k+1)-1)}{3}$

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 n

Question:8 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ :

Answer:

Let the given statement be p(n) i.e.

For n = 1 we have , which is true

For n = k we have

, Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1.2+{2.2}^2+{3.2}^3+...+(k+1){.2}^{k+1}$ $=1.2+{2.2}^2+{3.2}^3+...+k{.2}^k+(k+1){.2}^{k+1}$

$=(k-1)2^{k+1}+2+(k+1){.2}^{k+1}(using(i))$
$=2^{k+1}(k-1+k+1)+2$
$=2^{k+1}(2k)+2$
$=k{.2}^{k+2}+2$
$=(k+1-1){.2}^{k+1+1}+2$

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 n

Question:9 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=1-\frac{1}{2^n}$

Answer:

Let the given statement be p(n) i.e.
$p(n):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=1-\frac{1}{2^n}$
For n = 1 we have
$p(1):\frac{1}{2}=1-\frac{1}{2^1}=1-\frac{1}{2}=\frac{1}{2}$ , which is true

For n = k we have

$p(k):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}=1-\frac{1}{2^k}-(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{k+1}}$$=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}+\frac{1}{2^{k+1}}$

$=1-\frac{1}{2^k}+\frac{1}{2^{k+1}}(using(i))$
$=1-\frac{1}{2^k}(1-\frac{1}{2})$
$=1-\frac{1}{2^k}\left(\frac{1}{2}\right)$
$=1-\frac{1}{2^{k+1}}$

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 n

Question:10 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

Answer:

Let the given statement be p(n) i.e.
$p(n):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$
For n = 1 we have
$p(1):\frac{1}{2.5}=\frac{1}{10}=\frac{1}{(6(1)+4)}=\frac{1}{10}$ , which is true

For n = k we have

$p(k):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3k-1)(3k+2)}=\frac{k}{(6k+4)}-(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3(k+1)-1)(3(k+1)+2)}$$=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3k-1)(3k+2)}+\frac{1}{(3(k+1)-1)(3(k+1)+2)}$

$=\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}(using(i))$
$=\frac{1}{3k+2}(\frac{k}{2}+\frac{1}{3k+5})$
$=\frac{1}{3k+2}\left(\frac{k(3k+5)+2}{2(3k+5)}\right)$
$=\frac{1}{3k+2}\left(\frac{3k^2+5k+2}{2(3k+5)}\right)$
$=\frac{1}{3k+2}\left(\frac{3k^2+3k+2k+2}{2(3k+5)}\right)$
$=\frac{1}{3k+2}\left(\frac{(3k+2)(k+1)}{2(3k+5)}\right)$
$=\frac{(k+1)}{6k+10}$
$=\frac{(k+1)}{6(k+1)+4}$

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 n

Question:11 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

Answer:

Let the given statement be p(n) i.e.
$p(n):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$
For n = 1 we have
$p(1):\frac{1}{1.2.3}=\frac{1}{6}=\frac{1(1+3)}{4(1+1)(1+2)}=\frac{4}{4.2.3}=\frac{1}{6}$ , which is true

For n = k we have

$p(k):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)}-(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{(k+1)(k+2)(k+3)}$$=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$

$=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}(using(i))$
$=\frac{1}{(k+1)(k+2)}(\frac{k(k+3)}{4}+\frac{1}{k+3})$
$=\frac{1}{(k+1)(k+2)}\left(\frac{k(k+3{)}^2+4}{4(k+3)}\right)$
$=\frac{1}{(k+1)(k+2)}\left(\frac{k(k^2+9+6k)+4}{4(k+3)}\right)$
$=\frac{1}{(k+1)(k+2)}\left(\frac{k^3+9k+6k^2+4}{4(k+3)}\right)$
$=\frac{1}{(k+1)(k+2)}\left(\frac{k^3+2k^2+k+8k+4k^2+4}{4(k+3)}\right)$
$=\frac{1}{(k+1)(k+2)}\left(\frac{k(k^2+2k+1)+4(k^2+2k+1)}{4(k+3)}\right)$
$=\frac{1}{(k+1)(k+2)}\left(\frac{(k+1{)}^2(k+4)}{4(k+3)}\right)$
$=\frac{(k+1)((k+1)+3)}{4(k+1+1)(k+1+2)}$

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 n

Question:12 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $a+ar+a{r}^2+...+a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$

Answer:

Let the given statement be p(n) i.e.
$p(n):a+ar+a{r}^2+...+a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$
For n = 1 we have
$p(1):a=\frac{a(r^1-1)}{r-1}=\frac{r-1}{r-1}=1$ , which is true

For n = k we have

$p(k):a+ar+a{r}^2+...+a{r}^{k-1}=\frac{a(r^k-1)}{r-1}-(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):a+ar+a{r}^2+...+a{r}^k$$=a+ar+a{r}^2+...+a{r}^{k-1}+a{r}^k$

$=a.\frac{r^k-1}{r-1}+a{r}^k(using(i))$
$=\frac{a(r^k-1)+(r-1)a{r}^k}{r-1}$
$=\frac{a{r}^k(1+r-1)-a}{r-1}$
$=\frac{a{r}^k.r-a}{r-1}$
$=\frac{a(r^{k+1}-1)}{r-1}$

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 n

Question:13 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})..(1+\frac{(2n+1)}{n^2})=(n+1{)}^2$

Answer:

Let the given statement be p(n) i.e.
$p(n):(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})..(1+\frac{(2n+1)}{n^2})=(n+1{)}^2$
For n = 1 we have
$p(1):(1+\frac{3}{1})=4=(1+1{)}^2=2^2=4$ , which is true

For n = k we have

$p(k):(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})..(1+\frac{(2k+1)}{k^2})=(k+1{)}^2-(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})..(1+\frac{(2(k+1)+1)}{(k+1{)}^2})$ $=(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})..(1+\frac{2k+1}{k^2})(1+\frac{(2(k+1)+1)}{(k+1{)}^2})$

$=(k+1{)}^2(1+\frac{(2(k+1)+1)}{(k+1{)}^2})(using(i))$
$=(k+1{)}^2\left(\frac{(k+1{)}^2+(2(k+1)+1)}{(k+1{)}^2}\right)$
$=(k^2+1+2k+2k+2+1)$
$=(k^2+4k+4)$
$=(k+2{)}^2$
$=(k+1+1{)}^2$

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 n

Question:14 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{n})=(n+1)$