Principle of Mathematical Induction

Class 11 NCERT

NCERT

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

Solution :

Let the given statement be $P(n),$ i.e.,$\\$ $P(n):1+3+3^2+...+3^{n-1}=\dfrac{(3^n-1)}{2}$$\\$ For $n=1,$ we have $\\$ $P(1):=\dfrac{(3^1-1)}{2}=\dfrac{3-1}{2}\\ =\dfrac{2}{2}=1,$ which is true.$\\$ Let $P(k)$ be true for some positive integer k, i.e., $\\$ $1+3+3^2+...+3^{k-1} =\dfrac{3^k-1}{2}....(i)$$\\$ We shall now prove that $P(k+1)$ is true .$\\$ Consider $\\$ $1+3+3^2+....+3^{k-1}+3^{(k+1)-1}\\ =(1+3+3^2+....+3^{k-1})+3^k\\ =\dfrac{(3^k-1)}{2}+3^k \ \ \ \ \ \ $[Using(i)]$\\$

$=\dfrac{(3^k-1)+2.3^k}{2}\\ =\dfrac{(1+2)3^k-1}{ 2}\\ =\dfrac{3.3^k-1}{2}\\ =\dfrac{3^{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., $N.$

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

Solution :

Let the given statement be $P(n),$ i.e.,$\\$ $P(n):1^3+2^3+3^3+....+n^3=(\dfrac{n(n+1)}{2})^2$$\\$ For$n=1,$ we have $\\$ $P(1):1^3=1=(\dfrac{1(1+1)^2}{2})=(\dfrac{1.2}{2})^2\\ =1^2=1,$ which is true.$\\$ Let $P(k)$ be true for some positive integer k,i.e.,$\\$ $1^3+2^3+3^3+....+k^3=(\dfrac{k(k+1)}{2})^2...(i)$$\\$ We shall now prove that $P(k+1)$ is true .$\\$ Consider $\\$ $1^3+2^3+3^3+...+k^3+(k+1)^3\\ =(1^3+2^3+3^3+....+k^3)+(k+1)^3\\ =(\dfrac{k(k+1)}{2})^2 +(k+1)^3 \ \ \ \ \ [Using (i)]\\ =\dfrac{k^2(k+1)^2}{4}+(k+1)^3\\ =\dfrac{k^2(k+1)^2+4(k+1)^3}{4}\\ =\dfrac{(k+1)^2\{k^2+4(k+1)\}}{4}\\ =\dfrac{(k+1)^2(k+2)^2}{4}\\ =\dfrac{(k+1)^2(k+1+1)^2}{4}\\ =(\dfrac{(k+1)(k+1+1)}{2})^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.

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

Solution :

Let the given statement be $P(n),$ i. e.,$\\$ $P(n):1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...\\ .+\dfrac{1}{1+2+3+...n}=\dfrac{2n}{n+1}$$\\$ For $ n=1,$ we have $\\$ $ P(1):1=\dfrac{2.1}{1+1}=\dfrac{2}{2}=1,$ which is true.$\\$ Let $ P(k)$ be true for some positive integer k,i.e.,$\\$ $1+\dfrac{1}{1+2}+...+\dfrac{1}{1+2+3}+\\ ...+\dfrac{1}{1+2+3+...+k}=\dfrac{2k}{k+1}....(i)$$\\$ We shall now prove that $P(k+1)$ is true.$\\$ Consider $\\$ $1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\\ ....+\dfrac{1}{1+2+3+....+k}+\dfrac{1}{1+2+3+....+k+(k+1)}\\ =(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...k})\\ +\dfrac{1}{1+2+3+....+k+(k+1)}\\ =\dfrac{2k}{k+1}+\dfrac{1}{1+2+3+...+k+(k+1)} \ \ \ \ \ [Using (i)]\\ =\dfrac{2k}{k+1}+\dfrac{2}{(k+1)(k+1+1)}\\ [1+2+3+...+n=\dfrac{n(n+1)}{2}]\\ =\dfrac{2k}{(k+1)}+\dfrac{2}{(k+1)(k+2)}\\ =\dfrac{2}{(k+1)}(k+\dfrac{1}{k+2})\\ =\dfrac{2}{(k+1)}(\dfrac{k^2+2k+1}{k+2})\\ =\dfrac{2}{(k+1)}[\dfrac{(k+1)^2}{k+2}]\\ =\dfrac{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 i.e., N.

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

Solution :

Let the given statement be $P(n),$ i.e.,$\\$ $P(n):1.2.3+2.3.4+....+n(n+1)(n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$$\\$ For $ n=1 $ we have$\\$ $P(1):1.2.3=6=\dfrac{1(1+1)(1+2)(1+3)}{4}\\ =\dfrac{1.2.3.4}{4}=6,$ which is true.$\\$ Let$ P(k)$ be true for some positive integer k, i.e.,$\\$ $1.2.3+2.3.4+...+k(k+1)\\ (k+2)=\dfrac{k(k+1)(k+2)(k+3)}{4}....(i)$$\\$ 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)\\ =\dfrac{k(k+1)(k+2)(k+3)}{4} \\ +(k+1)(k+2)(k+3) \ \ \ \ \ [Using(i)] \\ -(k+1)(k+2)(k+3)(\dfrac{k}{4}+1)\\ =\dfrac{(k+1)(k+2)(k+3)(k+4)}{4}\\ =\dfrac{(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.

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

Solution :

Let the given statement be $P(n)$,i.e.,$\\$ $P(n):1.3+2.3^2+3.3^3+....+n3^n=\dfrac{(2n-1)3^{n+1}+3}{4}$$\\$ $P(1):1.3=3=\dfrac{(2.1-1)3^{1+1}+3}{4}\\ =\dfrac{3^2+3}{4}=\dfrac{12}{4}=3,$ which is true.$\\$ Let$ P(k) $be true for some positive integer k, i.e.,$\\$ $1.3+2.3^2+3.3^3+...+k3^k=\dfrac{(2k-1)3^{k+1}+3}{4}...(i)$$\\$ We shall now prove that $P(k+1)$ is true.$\\$

Consider $\\$ $1.3+2.3^2+3.3^3+....+k.3^k+(k+1).3^{k+1}\\ =(1.3+2.3^2+3.3^3+....+k.3^k)+(k+1).3^{k+1}\\ =\dfrac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k-1} \ \ \ \ [Using (i)]\\ =\dfrac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}\\ =\dfrac{3^{k+1}\{2k-1+4(k+1)\}+3}{4}\\ =\dfrac{3^{k+1}\{6k+3\}+3}{4}\\ =\dfrac{3^{k+1}.3\{2k+1\}+3}{4}\\ =\dfrac{3^{(k+1)+1}\{2k+1\}+3}{4}\\ =\dfrac{\{2(k+1)-1\}3^{(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.

Consider $\\$ $1.3+2.3^2+3.3^3+...+k.3^k+(k+1).3^{k+1}\\ =(1.3+2.3^2+3.3^3+...+k.3^k)+(k+1).3^{k+1}$

6   Prove that following by using the principle of mathematical induction for all $n \in N $:$\\$ $ 1+3+3^2+.....+3^{n-1}=\dfrac{(3^n-1)}{2}$

Solution :

Let the given statement be $P(n),$ i.e.,$\\$ $P(n):1+3+3^2+....+3^{n-1}=\dfrac{(3^n-1)}{2}$$\\$ For $ n=1,$ we have $\\$ $ P(1):=\dfrac{(3^1-1)}{2}=\dfrac{3-1}{2}=\dfrac{2}{2}=1,$ which is true$\\$ Let $ P(k)$ be true for some positive integer $k$, i.e.,$\\$ $ 1+3+3^2+.....+3^{k-1}=\dfrac{(3^k-1)}{2} .........(i)$$\\$ We shall now prove that $P( k + 1 )$ is true.$\\$ Consider$\\$ $1+3+3^2+.....+3^{k-1}+3^{(k+1)-1}\\ =(1+3+3^2+....+3^{k-1})+3^k\\ =\dfrac{(3^k-1)}{2}+3^k \qquad [\text{Using (i)}]\\ =\dfrac{(3^k-1)+2.3^k}{2}\\ =\dfrac{(1+2)3^k-1}{2}\\ =\dfrac{3.3^k-1}{2}\\ =\dfrac{3^{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., $N.$

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

Solution :

Let the given statement be $P ( n ) $, i.e.,$\\$ $P(n):1^3+2^3+3^3+......+n^3 =(\dfrac{n(n+1)}{2})^2$$\\$ For $ n=1, $ we have $\\$ $P(1):1^3=1=(\dfrac{1(1+1)^2}{2}) =(\dfrac{1.2}{2})^2=1^2=1$ which is true.$\\$ Let $P(k)$ be true for some positive integer $k$, i.e.,$\\$ $1^3+2^3+3^3+.....+k^3=(\dfrac{k(k+1)}{2})^2 ........(i)$$\\$ We shall now prove that $P (k + 1 )$ is true.$\\$ Consider $\\$ $1^3+2^3+3^3+.....+k^3+(k+1)^3\\ =(1^3+2^3+3^3+....+k^3)+(k+1)^3\\ =(\dfrac{k(k+1)}{2})^2+(k+1)^3 \quad [\text{Using (i)}]\\ =\dfrac{k^2(k+1)^2}{4}+(k+1)^3\\ =\dfrac{k^2(k+1)^2+4(k+1)^3}{4}\\ =\dfrac{(k+1)^2\{k^2+4k+4\}}{4}\\ =\dfrac{(k+1)^2(k+2)^2}{4}\\ =\dfrac{(k+1)^2(k+1+1)^2}{4}\\ =(\dfrac{(k+1)(k+1+1)}{2})^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., $N.$

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

Solution :

Let the given statement be $P( n) $ , i.e.,$\\$ $P(n):1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+.....+\dfrac{1}{(1+2+3+....n)}=\dfrac{2n}{n+1}$$\\$ For $n=1, $ we have$\\$ $P(1):1=\dfrac{2.1}{1+1}=\dfrac{2}{2}=1, $ which is true.$\\$ Let $P (k) $ be true for some positive integer $k$, i.e.,$\\$ $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+.....+\dfrac{1}{(1+2+3+....k)}=\dfrac{2k}{k+1}$$\\$ We shall now prove that $P(k+1)$ is true $\\$ Consider $\\$ $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+.....+\dfrac{1}{(1+2+3+....k)}+\dfrac{1}{(1+2+3+....k+(k+1))}\\ =(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+....+\dfrac{1}{(1+2+3+...k)})+\dfrac{1}{(1+2+3+....k+(k+1))}\\ =\dfrac{2k}{k+1}+\dfrac{1}{1+2+3+....+k+(k+1)} \quad [\text{Using (i)}]\\ =\dfrac{2k}{k+1}+\dfrac{1}{\dfrac{(k+1)(k+1+1)}{2}} \quad [1+2+3+....+n=\dfrac{n(n+1)}{2}]\\ =\dfrac{2k}{(k+1)}+\dfrac{2}{(k+1)(k+1+1)}\\ =\dfrac{2}{(k+1)}(k+\dfrac{1}{k+2})\\ =\dfrac{2}{(k+1)}(\dfrac{k^2+2k+1}{k+2})\\ =\dfrac{2}{(k+1)}[\dfrac{(k+1)^2}{k+2}]\\ =\dfrac{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 i.e., $N.$

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

Solution :

Let the given statement be $P(n)$, i.e.,$\\$ $P(n):1.2.3+2.3.4+.....+n(n+1)(n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$ For $n=1, $ we have $\\$ $P(1):1.2.3=6=\dfrac{1(1+1)(1+2)(1+3)}{4}=\dfrac{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+....+k(k+1)(k+2)=\dfrac{k(k+1)(k+2)(k+3)}{4}.....(i)$$\\$ We shall now prove that $P (k + 1 )$ is true.$\\$ Consider $\\$ $1.2.3+2.3.4+.....+k(k+1)(k+2)+(k+1)(k+2)(k+3)\\ =\{1.2.3+2.3.4+....+k(k+1)(k+2)\}+(k+1)(k+2)+(k+3)\\ =\dfrac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3) \quad [\text{Using(i)}]\\ =(k+1)(k+2)(k+3)(\dfrac{k}{4}+1)\\ =\dfrac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}$$\\$ 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.$

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

Solution :

Let the given statement be $P(n)$ , i.e.,$\\$ $P(n):1.3+2.3^2+3.3^3+.....+n3^n=\dfrac{(2n-1)3^{n+1}+3}{4}$$\\$ For $ n=1,$ we have $\\$ $P(1):1.3=3=\dfrac{(2.1-1)3^{1+1}+3}{4}=\dfrac{3^2+3}{4}=\dfrac{12}{4}=3,$ which is true.$\\$ Let $P(k) $ be true for some positive integer $ k $ , i.e.,$\\$ $1.3+2.3^2+3.3^3+....+k3^k=\dfrac{(2k-1)3^{k+1}+3}{4} ..........(i)$$\\$ We shall now prove that $P(k+1)$ is true.$\\$ Consider$\\$ $1.3+2.3^2+3.3^3+....+k.3^k+(k+1).3^{k+1}\\ =(1.3+2.3^2+3.3^3+....+k.3^k)+(k+1).3^{k+1}\\ =\dfrac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k-1} \quad [\text{Using(i)}]\\ =\dfrac{(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}\\ =\dfrac{3^{k+1}\{2k-1+4(k+1)\}+3}{4}\\ =\dfrac{3^{k+1}\{6k+3\}+3}{4}\\ =\dfrac{3^{k+1}.3\{2k+1\}+3}{4}\\ =\dfrac{3^{(k+1)+1}.\{2k+1\}+3}{4}\\ =\dfrac{\{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 i.e.,$ N.$

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

Solution :

Let the given statement be $ p(n),$ 1.e.,$\\$ $P(n): \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+......+\dfrac{1}{n(n+1)(n+2)}=\dfrac{n(n+3)}{4(n+1)(n+2)}$ $\\$ For $ n=1,$ we have $\\$ $P(1): \dfrac{1}{1.2.3}+\dfrac{1(1+.3}{4(1+1)(1+2)}=\dfrac{1.4}{4.2.3}=\dfrac{1}{1.2.3}$which is true.$\\$ Let P(k) be true for some positive integer k, i.e.,$\\$ $\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+......+\dfrac{1}{k(k+1)(k+2)}=\dfrac{k(k+3)}{4(k+1)(k+2)}.........(i)$ $\\$ We shall now prove that P ( k + 1 ) is true.$\\$ Consider$\\$ $[\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+......+\dfrac{1}{k(k+1)(k+2)}]+\dfrac{1}{k(k+1)(k+2)(k+3)}=\dfrac{k(k+3)}{4(k+1)(k+2)}+\dfrac{1}{k(k+1)(k+2)(k+3)}\quad [Using \ (i)]$$\\$ $\dfrac{1}{k(k+1)(k+2)}{\dfrac{k(k+3)}{4}+\dfrac{1}{k+3}}\\ =\dfrac{1}{k(k+1)(k+2)}{\dfrac{k(k+3)^2}{4(k+3)}}\\ =\dfrac{1}{k(k+1)(k+2)}{\dfrac{k(k^2+6k+9)+4}{4(k+3)}}\\ =\dfrac{1}{k(k+1)(k+2)}{\dfrac{k^3+6k^2+9k+4}{4(k+3)}}\\ =\dfrac{1}{k(k+1)(k+2)}{\dfrac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}}\\ =\dfrac{1}{k(k+1)(k+2)}{\dfrac{k(k^2+2k+1)+4(k^2+2k+1}{4(k+3)}}\\ =\dfrac{1}{k(k+1)(k+2)}{\dfrac{k(k+1)^2+4(k+1)^2}{4(k+3)}}\\ =\dfrac{(k+1)^2(k+4)}{4(k+1)(k+2)(k+3)}\\ =\dfrac{(k+1){(k+1)+3}}{4{(k+1)+1}{(k+1)+2}}$ $\\$ 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 i.e., N.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ $P(n) :a+ar+ar^2+....+ar^{n-1}=\dfrac{a(r^n-1)}{r-1}$$\\$ For $n=1,$ we have$\\$ $P(1):a=\dfrac{a(r^1-1)}{(r-1)}=a,$ which is true.$\\$ Let P(k) be true for some positive integer k, i.e.,$\\$ $a+ar+ar^2+....+ar^{k-1}=\dfrac{a(r^k-1)}{k-1}.......(i)$$\\$ We shall now prove that $ P ( k + 1 )$ is true.$\\$ Consider $\\$ ${a+ar+ar^2+.....+ar^{k-1}}+ar^{(k+1)-1}\\ =\dfrac{a(r^k-1)}{r-1}+ar^k \quad [\text{Using(i)}]\\ =\dfrac{a(r^k-1)+ar^k(r-1)}{r-1}\\ =\dfrac{a(r^k-1)+ar^{k-1}-ar^k}{r-1}\\ =\dfrac{ar^k-a+ar^{k+1}-ar^k}{r-1}\\ =\dfrac{ar^{k+1}-a}{r-1}\\ =\dfrac{a(r^{k+1}-1)}{r-1}$ 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 i.e., N.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ P(n):$(1+\dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9}).......(1+\dfrac{(2n+1)}{n^2})=(n+1)^2$ $\\$ For $ n=1,$ we have $\\$ P(1):$(1+\dfrac{3}{1})=4=(1+1)^2=2^2=4,$ which is true.$\\$ $(1+\dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9}).......(1+\dfrac{(2k+1)}{k^2})=(k+1)^2.......(1)$$\\$ We shall now prove that P ( k + 1 ) is true.$\\$ Consider$\\$ $[(1+\dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9}).......(1+\dfrac{(2k+1)}{k^2})]{1+\dfrac{2(k+1)+1}{(k+1)^2}}$ $\\$ $=(k+1)^2(1+\dfrac{2(k+1)+1}{(k+1)^2}) \quad [\text{Using (1)}] \\ =(k+1)^2[\dfrac{(k+1)^2+2(k+1)+1}{(k+1)^2}]\\ =(k+1)^2+2(k+1)+1\\ ={(k+1)+1}^2$ $\\$ 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.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ P(n):$(1+\dfrac{1}{1})(1+\dfrac{1}{2})(1+\dfrac{1}{3})......(1+\dfrac{1}{n})=(n+1)$$\\$ For $ n=1,$ we have $\\$ P(1):$(1+\dfrac{1}{1})=2=(1+1),$ which is true.$\\$ Let P(k):$(1+\dfrac{1}{1})(1+\dfrac{1}{2})(1+\dfrac{1}{3})......(1+\dfrac{1}{k})=(k+1)$$\\$ We shall now prove that P ( k + 1 ) is true.$\\$ Consider$\\$ $[(1+\dfrac{1}{1})(1+\dfrac{1}{2})(1+\dfrac{1}{3})......(1+\dfrac{1}{k})](1+\dfrac{1}{k+1})\\ =(k+1)(1+\dfrac{1}{k+1}) \quad [\text{Using(1)}]\\ =(k+1)[\dfrac{(k+1)+1}{(k+1)}]\\ =(k+1)+1$ $\\$ 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 i.e., N.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ P(n):$1^2+3^2+5^2+....+(2n-1)^2=\dfrac{n(2n-1)(2n+1)}{3}$ $\\$ For $ n=1,$ we have $\\$ P(1)=$1^2=1=\dfrac{1(2.1-1)(2.1+1)}{3}=\dfrac{1.1.3}{3}=1, $ which is true.$\\$ Let P(k) be true for some positive integer k, i.e.,$\\$ P(k)=$1^2+362+5^2+....+(2k-1)^2=\dfrac{k(2k-1)(2k+1)}{3}....(1)$$\\$ we shall now prove that $ P(k+1)$ is true.$\\$ Consider$\\$ ${1^2+3^2+5^2+...+(2k-1)^2}+{2(k+1)-1}^2\\ =\dfrac{k(2k-1)(2k+1)}{3}+(2k+2-1)^2 \quad [\text{Using(1)}]\\ =\dfrac{k(2k-1)(2k+1)}{3}+(2k+1)^2\\ =\dfrac{2(2k-1)(2k+1)+3(2k+1)^2}{3}\\ =\dfrac{(2k+1){k(2k-1)+3(2k+1)}}{3}\\ =\dfrac{(2k+1){2k^2-k+6k+3}}{3}\\ =\dfrac{(2k+1){2k^2+5k+3}}{3}\\ =\dfrac{(2k+1){2k^2+2k+3k+3}}{3}\\ =\dfrac{(2k+1){2k(k+1)+3(k+1)}}{3}\\ =\dfrac{(2k+1)(k+1)(2k+3)}{3}\\ =\dfrac{(k+1){2(k+1)-1}{2(k+1)+1}}{3}$ $\\$ 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 i.e., N.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ P(n):$\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$ $\\$ For $ n=1,$ we have $\\$ P(1)=$ \dfrac{1}{1.4}=\dfrac{1}{3.1+1}=\dfrac{1}{4}+\dfrac{1}{1.4},$ which is true.$\\$ Let P(k) be true for some positive integer k, i.e.,$\\$ P(k):$\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{(3k-2)(3k+1)}=\dfrac{k}{(3k+1)}$ $\\$ We shall now prove that P ( k + 1 ) is true. $\\$ Consider $\\$ ${$\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{(3k-2)(3k+1)}}+\dfrac{1}{{3(k+1)-2}{3(k+1)+1}}\\ =\dfrac{k}{3k+1}+\dfrac{1}{(3k+1)(3k+4)} \quad [\text{Using(1)}]\\ =\dfrac{1}{3k+1}{k+\dfrac{1}{(3k+4)}}\\ =\dfrac{1}{3k+1}{\dfrac{k(3k+4)+1}{(3k+4)}}\\ =\dfrac{1}{(3k+1)}{\dfrac{3k^2+4k+1}{(3k+4)}}\\ =\dfrac{1}{(3k+1)}{\dfrac{3k^2+3k+k+1}{(3k+4)}}\\ =\dfrac{(3k+1)(k+1)}{(3k+1)(3k+4)}\\ =\dfrac{(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 i.e., N.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ P(n):$\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+.....+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n}{3(2n+3)}$ $\\$ For $ n=1,$ we have $\\$ P(1):$\dfrac{1}{3.5}=\dfrac{1}{3(2.1+3)}=\dfrac{1}{3.5},$ which is true.$\\$ Let P(k) be true for some positive integer k, i.e.,$\\$ $[\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+.....+\dfrac{1}{(2n+1)(2n+3)}]+\dfrac{1}{{2(k+1)+1}{2(k+1)+3}}\\ =\dfrac{k}{3(2k+3)}+\dfrac{1}{(2k+3)(2k+5)} \quad [\text{Using (1)}]\\ =\dfrac{1}{(2k+3)}[\dfrac{k}{3}+\dfrac{1}{(2k+5)}]\\ =\dfrac{1}{(2k+3)}[\dfrac{k(2k+5)+3}{3(2k+5)}]\\ =\dfrac{1}{(2k+3)}[\dfrac{2k^2+5k+3}{3(2k+5)}]\\ =\dfrac{1}{(2k+3)}[\dfrac{2k^2+2k+3k+3}{3(2k+5)}]\\ =\dfrac{1}{(2k+3)}[\dfrac{2k(k+1)+3(k+1}{3(2k+5)}]\\ =\dfrac{(k+1)}{3{2(k+1)+3}}$ $\\$ 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 i.e., N.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ P(n):$ 1+2+3+....+n < \dfrac{1}{8}(2n+1)^2$ $\\$ It can be noted that P(n) is true for $n = 1$ since $\\$ $1 < \dfrac{1}{8}(2.1+1)^2=\dfrac{9}{8}$ $\\$ Let P(k) be true for some positive integer k, i.e.,$\\$ $1+2+....+k < \dfrac{1}{8}(2k+1)^2+(k+1) \quad [\text{Using (1)}]\\ < \dfrac{1}{8}{(2k+1)^2+8(k+1)}\\ < \dfrac{1}{8}{4k^2+4k+1+8k+8}\\ < \dfrac{1}{8}{4k^2+12k+9}\\ < \dfrac{1}{8}(2k+3)^2\\ < \dfrac{1}{8}{2(k+1)+1}^2$ $\\$ Hence, $ (1+2+3+...+k)+(k+1) < \dfrac{1}{8}(2k+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 i.e., N.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ P ( n ) : $n ( n + 1 )( n + 5 )$ , which is a multiple of 3.$\\$ It can be noted that P ( n ) is true for n = 1 since$ 1 ( 1 + 1 )( 1 + 5 )= 12$ , which is a multiple of 3.$\\$ Let P(k) be true for some positive integer k, i.e.,$\\$ $k ( k + 1 )( k + 5 )$ is a multiple of 3.$\\$ $\therefore k ( k + 1 )( k + 5)= 3 m$, where $m\in N .... ( 1 )$$\\$ We shall now prove that P ( k + 1 ) is true whenever P(k) is true.$\\$ Consider$\\$ $(k+1){(k+1)+1}{(k+1)+5}\\ =(k+1)(k+2){(k+5)+1}\\ =(k+1)(k+2)(k+5)+(k+1)(k+2)\\ ={k(k+1)(k+5)+2(k+1)(k+5)}+(k+1)(k+2)\\ =3m+(k+1){2(k+5)+(k+2)}\\ =3m+(k+1){2k+10+k+2}\\ =3m+(k+1){3k+12}\\ =3m+3(k+1){k+4}\\ =3{m+(k+1)(k+4)}=3*q,$ where $ q={m+(k+1)(k+4)}$ is some natural number. $\\$ Therefore,$(k+1){(k+1)+1}{(k+1)+5}$ is a multiple of 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.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ P ( n ) :$10^( 2 n - 1 )+ 1$ is divisible by 11.$\\$ It can be observed that P ( n ) is true for $n = 1$$\\$ Since 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 k- 1 }+ 1$ is divisible by 11.$\\$ $\therefore 10^{ 2 k - 1 }+ 1 = 11 m ,$ where$\\$ $m\in N .... ( 1 )$ $\\$ We shall now prove that P ( k + 1 ) is true whenever P(k) is true.$\\$ Consider$\\$ $10^{2(k+1)-1}+1\\ =10^{2k+2-1}+1\\ =10^{2k+1}+1\\ =10^{2}(10^{2k-1}+1-1)+1\\ =10^2(10^{2k-1}+1)-10^2+1\\ =10^2.11m-100+1\quad [\text{Using(1)}]\\ =100*11m-99\\ =11(100m-9)\\ =11r,$ where $ r=(100m-9)$ is some natural number$\\$ Therefore,$10^{2(k+1)-1}+1$ 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.

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

Solution :

Let 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 $x^{2 * 1} - y ^{2* 1 }= x^ 2 - y ^2 =( x + y )( x - y )$ is divisible by $( x + y ) .$$\\$ Let P ( k ) be true for some positive integer k, i.e.,$\\$ $x^ {2 k }- y^{ 2 k}$ is divisible by $x + y$ .$\\$ $\therefore$ Let $x ^{2 k} - y ^{2 k }= m ( x + y )$ , where $m\in N ..... ( 1 )$$\\$ We shall now prove that P ( k + 1 ) is true whenever P(k) is true.$\\$ Consider$\\$ $x^{2(k+1)}-y^{2(k+1)}\\ =x^{2k}.x^2-y^{2k}+y^2\\ =x^2(x^{2k}-y^{2k}+y^{2k})-y^{2k}.y^2\\ =x^2{m(x+y)+y^{2k}}-y^{2k}.y^2\quad [\text{Using (1)}] \\ =m(x+y)x^2+y^{2k}.x^2-y^{2k}.y^2\\ =m(x+y)x^2+y^{2k}(x^2-y^2)\\ =m(x+y)x^2+y^{2k}(x+y)(x-y)\\ =(x+y){mx^2+y^{2k}(x-y)},$ 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.

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

Solution :

Let the given statement be P(n), i.e.,$\\$ P ( n ) :$3^{ 2 n + 2} - 8 n - 9$ is divisible by 8.$\\$ It can be observed that P(n) is true for n = 1$\\$ Since $3 ^{2 * 1 + 2 }- 8 * 1 - 9 = 64$, which is divisible by 8.$\\$ Let P(k) be true for some positive integer$\\$ k , i.e., $3^{ 2 k + 2 }- 8 k - 9$ is divisible by 8.$\\$ $\therefore 3 ^{2 k + 2 }- 8 k - 9 = 8 m $; where $m \in N ....... ( 1 )$$\\$ We shall now prove that P ( k + 1 ) is true whenever P ( k ) is true.$\\$ Consider$\\$ $3^{2(k+1)+2}-8(k+1)-9\\ 3^{2k+2}.3^2-8k-8-9\\ =3^2(3^{2k+2}-8k-9+8k+9)-8k-17\\ =3^2(3^{2k+2}-8k-9)+3^2(8k+9)-8k-17\\ =9.8m+9(8k+9)-8k-17\\ =9.8m+72k+81-8k-17\\ =9.8m+64k+64\\ =8(9m+8k+8)\\ =8r,$ where $ r=(9m+8k+8)$ is a natural number$\\$ Therefore, $ 3^{2(k+1)+2}-8(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 numbers i.e., N.

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

Solution :

Let the given statement be P ( n ) , i.e.,$\\$ P ( n ) :$41^ n - 14 ^n $ is a multiple of 27.$\\$ It can be observed that P ( n ) is true for $n = 1$ $\\$ Since $ 41 ^1 - 14^ 1 = 27 $ , which is a multiple of 27.$\\$ Let P(k) be true for some positive integer k, i.e.,$\\$ $\therefore 41^ k - 14^ k$ is a multiple of 27$\\$ $\therefore 41 ^k - 14^ k = 27 m , m\in N ...... ( 1 )$$\\$ We shall now prove that P ( k + 1 ) is true whenever P(k) is true.$\\$ Consider$\\$ $41^{k+1}-14^{k+1}\\ =41^ k .41 - 14^ k .14\\ = 41 ( 41^ k - 14^ k + 14^ k )- 14 ^k . 14\\ = 41.27 m + 14^ k ( 41 - 14 )\\ = 41.27 m + 27.14 ^k\\ = 27 ( 41 m -14^ k )\\ = 27 * r$ , where$ r =( 41 m - 14^ k )$ is a natural number.$\\$ Therefore,$ 41^{ k + 1 }-14^{ k + 1}$ is a multiple of 27.$\\$ 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.

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

Solution :

Let the given statement be P ( n ) , i.e.,$\\$ P ( n ) :$ ( 2 n + 7 ) < ( n + 3 )^2 $ $\\$ It can be observed that P ( n ) is true for $ n = 1$ $\\$ Since $2.1 + 7 = 9 <( 1+ 3 )^2= 16$ , which is true. $\\$ Let P ( k ) be true for some positive integer k, i.e., $\\$ $( 2 k + 7) < ( k + 3 )^ 2$ $\\$ We shall now prove that P ( k + 1 ) is true whenever P ( k ) is true. $\\$ Consider $\\$ $( 2 (k + 1 )+ 7 )= (2 k + 7 )+ 2\\ \therefore 2 ( k + 1 )+ 7 =( 2 k + 7 )+ 2 < ( k + 3 )^ 2+ 2 \quad [\text{Using (1)}] \\ 2 ( k + 1 )+ 7 < k ^2 + 6 k + 9 + 2\\ 2 ( k + 1 )+ 7 < k ^2 + 6 k + 11\\ $ $\\$ Now, $ k^2+6k+11 < k^2+8k+16\\ \therefore 2(k+1)+7 < (k+4)^2\\ 2(k+1)+7 < {(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 i.e., N.