# 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 $n \in N : 1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+....+\dfrac{1}{(1+2+3+...+n)}=\dfrac{2n}{n+1}$

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