(S\_{n} = \frac{1+2+3+...+n}{n} ) then (S\_{1}^{2} +S\_{2}^{... | Mathematics | SolveeIt | Solveeit

Today, August 24

Question

$S_{n} = \frac{1+2+3+...+n}{n}$ then $S_{1}^{2} +S_{2}^{2} + S^{2}_{3}+....+S^{2}_{ n} =$

A

$\frac{n}{24} \left(2n^{2} +9n+13\right)$

B

$\frac{1}{24} \left(2n^{2} +9n+13\right)$

C

$\frac{n^2}{24} \left(2n^{2} +9n+13\right)$

D

$\frac{n}{24} \left(2n^{2} - 9n+13\right)$

Answer

$\frac{n}{24} \left(2n^{2} +9n+13\right)$

Explanation

Solution

Put n = 2 and verify the options . $S_1^2 + S_2^2 = 1 + \frac{9}{4} = \frac{13}{4}$