Answer to Question #231396 in Discrete Mathematics for citra

Question

Provethat12 +32 +52 +⋯+(2n+1)2 = (n+1)(2n+1)(2n + 3)∕3 whenever n is a nonnegative integer.

Answer 1

Let us prove that $1^2 +3^2 +5^2 +⋯+(2n+1)^2 = \frac{(n+1)(2n+1)(2n + 3)}{3}$ using method of mathematical induction.

For $n=0$ the left side is equal to $1^1=1,$ and the right side is equal to $\frac{(0+1)(2\cdot 0+1)(2\cdot 0 + 3)}{3}\frac{3}{3}=1.$

Let for $n=k$ the formula $1^2 +3^2 +5^2 +⋯+(2k+1)^2 = \frac{(k+1)(2k+1)(2k + 3)}{3}$ is true.

Let us prove for $n=k+1:$

$1^2 +3^2 +5^2 +⋯+(2k+1)^2+(2(k+1)+1)^2 = \frac{(k+1)(2k+1)(2k + 3)}{3}+(2k+3)^2$

$= (2k+3)(\frac{(2k^2+3k+1}{3}+(2k+3))=(2k+3)\frac{2k^2+3k+1+6k+9}{3}=(2k+3)\frac{2k^2+9k+10}{3}$

$=(2k+3)\frac{(k+2)(2k+5)}{3}=\frac{(k+2)(2k+3)(2k+5)}{3}=\frac{((k+1)+1)(2(k+1)+1)(2(k+1)+3)}{3}.$

We conclude that this formula is true for any non-negative integer number $n.$