If 1² + 3² + 5² + ..... + (2n - 1)² = \(\rm \frac{n(2n-1)(2n+1)}{3}\), then 12 + 32 + 52 + ..... + (2n + 1)² equal :

Calculation:

Given, 12 + 32 + 52 + ..... + (2n - 1)2 = \(\rm \frac{n(2n-1)(2n+1)}{3}\)

It represents the sum of the first n odd numbers.

∴ 12 + 32 + 52 + ..... + (2n + 1)² 

=  12 + 32 + 52 + .... + (2n - 1)2  + (2n + 1)2 

= \(\rm \frac{n(2n-1)(2n+1)}{3}\) + (2n + 1)²

= \(\rm (2n+1)\left[ \frac{n(2n-1)}{3}+2n+1\right]\)

= \(\rm \frac{(2n+1)}{3}\left[ n(2n-1)+6n+3\right]\)

= \(\frac{(2n+1)}{3}\left[ 2n^2-n+6n+3\right]\)

= \(\rm\frac{(2n+1)}{3}\left[ 2n^2+5n+3\right]\)

= \(\rm\frac{(2n+1)}{3}\left[ 2n^2+2n+3n+3\right]\)

= \(\rm\frac{(2n+1)}{3}\left[ 2n(n+1)+3(n+1)\right]\)

= \(\rm \frac{(n+1)(2n+1)(2n+3)}{3}\)

∴  12 + 32 + 52 + ..... + (2n + 1)2  = \(\rm \frac{(n+1)(2n+1)(2n+3)}{3}\)