Question

Let $S_k=\frac{1+2+3+\cdots +k}{k}.$ If
$S_1^2+S_2^2+\cdots +S_{19}^2=\frac{A}{4},$ then the value of $A$ is

Answer 1

$S_k=\frac{\frac{k(k+1)}{2}}{k}=\frac{k+1}{2}$
$S_1^2+S_2^2+\cdots +S_{19}^2$
$=\frac{1}{4}(2^2+3^2+4^2+\cdots +{20}^2)$
$=\frac{1}{4}(1^2+2^2+3^2+4^2+\cdots +{20}^2-1)$
$=\frac{1}{4}[\frac{20\left(21\right)\left(41\right)}{6}-1]$
$=\frac{2869}{4}$
$\therefore A=2869$