# 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}$$\\ Thus,P(k+1) is true whenever P(k) is true.\\ Hence , by the principle of mathematical induction,statement P(n) is true for all natural numbers i.e., N. Consider \\ 1.3+2.3^2+3.3^3+...+k.3^k+(k+1).3^{k+1}\\ =(1.3+2.3^2+3.3^3+...+k.3^k)+(k+1).3^{k+1} 6 Prove that following by using the principle of mathematical induction for all n \in N :\\ 1+3+3^2+.....+3^{n-1}=\dfrac{(3^n-1)}{2} ##### Solution : Let the given statement be P(n), i.e.,\\ P(n):1+3+3^2+....+3^{n-1}=\dfrac{(3^n-1)}{2}$$\\$ 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.

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 $n \in N :$ $\\$ $\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{3n+1}$