Question

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

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$1^2+3^2+5^2+.......+({2n-1)}^2=\frac{n(2n-1)(2n+1)}{3}$
For $n=1$ we have
$P\left(1\right)=1^2=1=\frac{1(2.1-1)(2.1+1)}{3}=\frac{\mathrm{1.1.3}}{3}=1$, which is true.
Let $P\left(k\right)$ be true for some $k\in N$ i.e.,
$1^2+3^2+5^2+.......+({2n-1)}^2=\frac{k(2k-1)(2k+1)}{3}.....(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$1^2+3^2+5^2+.......+({2k-1)}^2+{\left\{2\right(k+1)-1\}}^2$ [Using $\left(i\right)$]
$=\frac{k(2k-1)(2k+1)}{3}+{(2k+2-1)}^2$
$=\frac{k(2k-1)(2k+1)}{3}+{(2k+1)}^2$
$=\frac{k(2k-1)(2k+1)+3{(2k+1)}^2}{3}$
$=\frac{(2k+1)\left\{k\right(2k-1)+3(2k+1\left)\right\}}{3}$
$=\frac{(2k+1)(2k^2-k+6k+3)}{3}$
$=\frac{(2k+1)(2k^2+5k+3)}{3}$
$=\frac{(2k+1)(2k^2+2k+3k+3)}{3}$
$=\frac{(2k+1)\left\{2k\right(k+1)+3(k+1\left)\right\}}{3}$
$=\frac{(2k+1)(k+1)(2k+3)}{3}$
$=\frac{(k+1)\left\{2\right(k+1)-1\}\left\{2\right(k+1)+1\}}{3}$
Thus $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$.