Question

$1+3+5+.....+(2n-1)=n^2$ i.e., the sum of first n odd natural numbers is $n^2$.

Answer 1

Let $P\left(n\right):1+3+5+........+(2n-1)=n^2$

For n = 1

P(1) : 1 = $1^2$

$1=1$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

(k) : $1+3+5+......+(2k-1)=k^2$ .....(1)

We have to show that

$1+3+5+......+(2k-1)+2(k+1)-1=(k+1{)}^2$

Now,

$1+3+5+...+(2k-1)+2(2k+1)$

$=k^2+(2k+1)$ [Using equation (1)]

$=k^2=2k+1$

$=(k+1{)}^2$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N by PMI