Question

Prove that the sum of the first n non-zero even numbers is $n^2+n.$

Answer 1

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Since the series is even, $2,4,6,8,.............,2n$

$a=2,d=2$

$\therefore S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

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

$=n[2+n-1]$

$=n(n+1)$

$\therefore S_n=n^2+n$