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
$p(k+1):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k+1)}$ $=\left ( 1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)} \right )+\frac{1}{(1+2+3+...+k+k+1)}$

$=\frac{2k}{k+1}+\frac{1}{(1+2+3+...+k+(k+1))} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{2k}{k+1}+\frac{1}{\frac{(k+1)(k+1+1)}{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because1+2+....+n = \frac{n(n+1)}{2} )$
$=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}$
$=\frac{2}{k+1}\left (k+\frac{1}{k+2} \right )$
$=\frac{2}{k+1}\left ( \frac{k^2+2k+1}{k+2} \right )$
$=\frac{2}{k+1}.\frac{(k+1)^2}{k+2}$
$=\frac{2(k+1)}{k+2}$
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}}$

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

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)}$

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

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)}$

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

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 $n\in\mathbb{N}$ : $a+ar+ar^2+...+ar^{n-1}=\frac{a(r^n-1)}{r-1}$

Answer:

Let the given statement be p(n) i.e.
$p(n):a+ar+ar^2+...+ar^{n-1}=\frac{a(r^n-1)}{r-1}$
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

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

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

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 $n\in\mathbb{N}$ : $\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2n+1)}{n^2} \right )=(n+1)^2$

Answer:

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

For n = k we have

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

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

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 $n\in\mathbb{N}$ : $\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{n} \right )=(n+1)$

Answer:

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

For n = k we have

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

$=(k+1)\left ( 1+\frac{1}{k+1} \right ) \ \ \ \ \ \ \ (using \ (i))$
$=(k+1)\left ( \frac{k+1+1}{k+1} \right )$
$=(k+2)$
$=(k+1+1)$
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:15 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}$

Answer:

Let the given statement be p(n) i.e.
$p(n):1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}$
For n = 1 we have
$p(1):1^2=1=\frac{1(2(1)-1)(2(1)+1)}{3}= \frac{1.1.3}{3}=1$ , which is true

For n = k we have

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

$=\frac{k(2k-1)(2k+1)}{3}+(2(k+1)-1)^2 \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k(2k-1)(2k+1)+3(2(k+1)-1)^2}{3}$
$=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}$
$=\frac{(2k+1)(k(2k-1)+3(2k+1))}{3}$
$=\frac{(2k+1)(2k^2-k+6k+3)}{3}$
$=\frac{(2k+1)(2k^2+5k+3)}{3}$
$=\frac{(2k+1)(2k^2+2k+3k+3)}{3}$
$=\frac{(2k+1)(2k+3)(k+1)}{3}$
$=\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$

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:16 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$

Answer:

Let the given statement be p(n) i.e.
$p(n):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$
For n = 1 we have
$p(1):\frac{1}{1.4}=\frac{1}{4}=\frac{1}{(3(1)+1)}=\frac{1}{3+1}=\frac{1}{4}$ , which is true

For n = k we have

$p(k):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}=\frac{k}{(3k+1)} \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3(k+1)-2)(3(k+1)+1)}$$=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}+\frac{1}{(3(k+1)-2)(3(k+1)+1)}$

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

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:17 Prove the following by using the principle of mathematical induction for all $n\in N$ : $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

Answer:

Let the given statement be p(n) i.e.
$p(n):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$
For n = 1 we have
$p(1):\frac{1}{3.5}= \frac{1}{15}=\frac{1}{3(2(1)+3)}=\frac{1}{3.5}=\frac{1}{15}$ , which is true

For n = k we have

$p(k):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)} \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2(k+1)+1)(2(k+1)+3)}$$=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}+\frac{1}{(2(k+1)+1)(2(k+1)+3)}$

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

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 n

Question:18 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $1+2+3+...+n<\frac{1}{8}(2n+1)^2.$

Answer:

Let the given statement be p(n) i.e.
$p(n):1+2+3+...+n<\frac{1}{8}(2n+1)^2.$
For n = 1 we have
$p(1):1<\frac{1}{8}(2(1)+1)^2= \frac{1}{8}(3)^2=\frac{9}{8}$ , which is true

For n = k we have

$p(k):1+2+3+...+k<\frac{1}{8}(2k+1)^2 \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this statement is true
Now,
For n = k + 1 we have
$p(k+1):1+2+3+...+k+1$$=1+2+3+...+k+k+1$

$< \frac{1}{8}\left ( 2k+1 \right )^2+(k+1) \ \ \ \ \ \ \ \ \ (using \ (i))$
$< \frac{1}{8}\left ( (2k+1)^2+8(k+1) \right )$
$< \frac{1}{8}\left ( 4k^2+4k+1+8k+8 \right )$
$< \frac{1}{8}\left ( 4k^2+12k+9\right )$
$< \frac{1}{8}\left ( 2k+3\right )^2$
$< \frac{1}{8}\left ( 2(k+1)+1\right )^2$

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:19 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $n(n+1)(n+5)$ is a multiple of $3.$

Answer:

Let the given statement be p(n) i.e.
$p(n):n(n+1)(n+5)$
For n = 1 we have
$p(1):1(1+1)(1+5)=1.2.6=12$ , which is multiple of 3, hence true

For n = k we have

$p(k):k(k+1)(k+5) \ \ \ \ \ \ \ -(i)$ , Let's assume that this is multiple of 3 = 3m
Now,
For n = k + 1 we have
$p(k+1):(k+1)((k+1)+1)((k+1)+5)$$=(k+1)(k+2)((k+5)+1)$

$=(k+1)(k+2)(k+5)+(k+1)(k+2)$
$=k(k+1)(k+5)+2(k+1)(k+5)+(k+1)(k+2)$
$=3m+2(k+1)(k+5)+(k+1)(k+2) \ \ \ \ \ \ \ \ (using \ (i))$
$=3m+(k+1)(2(k+5)+(k+2)) \ \ \ \ \ \ \ \ (using \ (i))$
$=3m+(k+1)(2k+10+k+2)$
$=3m+(k+1)(3k+12)$
$=3m+3(k+1)(k+4)$
$=3(m+(k+1)(k+4) )$
$=3l$ Where $\left ( l=(m+(k+1)(k+4) ) \right )$ some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is multiple of 3 for all natural numbers n

Question:20 Prove the following by using the principle of mathematical induction for all $n\in N$ : $10^{2n-1}+1$ is a divisible by $11.$

Answer:

Let the given statement be p(n) i.e.
$p(n):10^{2n-1}+1$
For n = 1 we have
$p(1):10^{2(1)-1}+1= 10^{2-1}+1=10^1+1=11$ , which is divisible by 11, hence true

For n = k we have

$p(k):10^{2k-1}+1 \ \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is divisible by 11 = 11m
Now,
For n = k + 1 we have
$p(k+1):10^{2(k+1)-1}+1$$=10^{2k+2-1}+1$
nbsp;
$=10^{2k+1}+1$
$=10^2(10^{2k-1}+1-1)+1$
$=10^2(10^{2k-1}+1)-10^2+1$
$=10^2(11m)-100+1 \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=100(11m)-99$
$=11(100m-9)$
$=11l$ Where $l= (100m-9)$ some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 11 for all natural numbers n

Question:21 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $x^2^n-y^2^n$ is divisible by $x+y.$

Answer:

Let the given statement be p(n) i.e.
$p(n):x^2^n-y^2^n$
For n = 1 we have
$p(1):x^{2(1)}-y^{2(1)}= x^2-y^2=(x-y)(x+y)$ , which is divisible by $(x+y)$ , hence true $(using \ a^2-b^2=(a+b)(a-b))$

For n = k we have

$p(k):x^{2k}-y^{2k} \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is divisible by $(x+y)$ $=(x+y)m$
Now,
For n = k + 1 we have
$p(k+1):x^{2(k+1)}-y^{2(k+1)}$ $=x^{2k}.x^2-y^{2k}.y^2$

$=x^2(x^{2k}+y^{2k}-y^{2k})-y^{2k}.y^2$
$=x^2(x^{2k}-y^{2k})+x^2.y^{2k}-y^{2k}.y^2$
$=x^2(x+y)m+(x^2-y^2)y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=x^2(x+y)m+((x-y)(x+y))y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ a^2-b^2=(a+b)(a-b))$
$=(x+y)\left ( x^2.m+(x-y).y^{2k} \right )$
$=(x+y)l$ where $l = (x^2.m+(x-y).y^{2k})$ some natural number

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

Question:22 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $3^{2n+2}-8n-9$ is divisible by $8.$

Answer:

Let the given statement be p(n) i.e.
$p(n):3^{2n+2}-8n-9$
For n = 1 we have
$p(1):3^{2(1)+2}-8(1)-9= 3^4-8-9=81-17=64=8\times 8$ , which is divisible by 8, hence true

For n = k we have

$p(k):3^{2k+2}-8k-9 \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is divisible by 8 = 8m
Now,
For n = k + 1 we have
$p(k+1):3^{2(k+1)+2}-8(k+1)-9$$=3^{2k+2+2}-8(k+1)-9$
$=3^{2k+2}.3^2-8k-8-9$
$=3^2(3^{2k+2}-8k-9+8k+9)-8k-17$
$=3^2(3^{2k+2}-8k-9)+3^2(8k+9)-8k-17$
$=9\times 8m+9(8k+9)-8k-17 \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=9\times 8m+72k+81-8k-17$
$=9\times 8m+80k-64$
$=9\times 8m+8(10k-8)$
$=8(9m+10k-8)$
$=8l$ where $l= 9m+10k-8$ some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 8 for all natural numbers n

Question:23 Prove the following by using the principle of mathematical induction for all $n\in\mathbb{N}$ : $41^n-14^n$ is a multiple of $27.$

Answer:

Let the given statement be p(n) i.e.
$p(n):41^n-14^n$
For n = 1 we have
$p(1):41^1-14^1= 41-14= 27$ , which is divisible by 27, hence true

For n = k we have

$p(k):41^k-14^k \ \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is divisible by 27 = 27m
Now,
For n = k + 1 we have
$p(k+1):41^{k+1}-14^{k+1}$$=41^{k}.41-14^{k}.14$
$=41(41^{k}-14^k+14^k)-14^{k}.14$
$=41(41^{k}-14^k)+14^k.41-14^{k}.14$
$=41(27m)+14^k(41-14) \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=41(27m)+14^k.27$
$=27(41m+14^k)$
$=27l$ where $l = 41m+14^k$ some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 27 for all natural numbers n

Question:24 Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ : $(2n+7)<(n+3)^2$

Answer:

Let the given statement be p(n) i.e.
$p(n):(2n+7)<(n+3)^2$
For n = 1 we have
$p(1):(2(1)+7)<(1+3)^2\Rightarrow 9< 16$ , which is true

For n = k we have

$p(k):(2k+7)<(k+3)^2 \ \ \ \ \ \ \ \ \ \ \ -(i)$ , Let's assume that this is true
Now,
For n = k + 1 we have
$p(k+1):(2(k+1)+7)$$=(2k+2+7)$
$<(k+3)^2+2 \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$<k^2+9+6k+2$
$<k^2+6k+11$
$Now , \ <k^2+6k+11<k^2+8k+16$
$<(k+4)^2$
$<((k+1)+3)^2$

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