Question

Using the principle of mathematical induction, prove the following for all $n\in N$:
$1^2+3^2+5^2+7^2+....+(2n-1{)}^2=\frac{n(2n-1)(2n+1)}{3}$

Answer 1

Let $P\left(k\right)$ be true for some $k\in N$ i.e.,

$P\left(k\right):1^2+3^2+5^2+....+(2k-1{)}^2=\frac{k(2k-1)(2k+1)}{3}$.

Now, $\{1^2+3^2+5^2+...+(2k-1{)}^2\}+{\left\{2\right(k+1)-1\}}^2$

$=\frac{k(2k-1)(2k+1)}{3}+(2k+1{)}^2=\frac{1}{3}.\left\{k\right(2k-1\left)\right(2k+1)+3(2k+1{)}^2\}$

$=\frac{1}{3}(2k+1)\left\{k\right(2k-1)+3(2k+1\left)\right\}=\frac{1}{3}(2k+1)(2k^2+5k+3)$

$=\frac{1}{3}(k+1)(2k+1)(2k+3)$.

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., $n.$