Question

find the sum to n terms of the series 1^2 + (1^2 + 2^2) + (1^2 + 2^2 +3^2) +...​

Answer 1

Given series :

$\sf 1^2 + (1^2 + 2^2) + (1^2 + 2^2 +3^2) +...n \: terms$

$\sf n \: term \:T_n = 1^2 + 2^2 +3^2+... + {n}^{2}$

$\sf T_n = \dfrac{n(n + 1)(2n + 1)}{6}$

$\sf T_n = \dfrac{1}{6} n(2 {n}^{2} + 3n + 1)$

$\sf T_n = \dfrac{1}{6} (2 {n}^{3} + 3{n}^{2} + n)$

$\sf T_n = \dfrac{ {n}^{3} }{3} + \dfrac{ {n}^{2} }{2} + \dfrac{n}{6}$

$\sf Sum \: of \: n \: terms = S_n = \sum \limits_1^n T_n$

$\sf S_n = \dfrac{1}{3} \dfrac{ {n}^{2} {(n + 1)}^{2} }{4} + \dfrac{1}{2} \dfrac{n(n + 1)}{6} + \dfrac{1}{6} \dfrac{n(n + 1)}{6}$

$\sf S_n = \dfrac{n(n + 1) }{12} [n(n + 1) + (2n + 1) + 1]$

$\sf S_n = \dfrac{n(n + 1) }{12} [ {n}^{2} + 3n + 2]$

$\sf S_n = \dfrac{n(n + 1) }{12} (n + 1)(n + 2)$

$\sf S_n = \dfrac{n(n + 1) ^{2} (n + 2) }{12}$