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Question:

What is the sum of n terms of the series $1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ....$

Posted by: Abhinay M. on 12.03.2018

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Answer:

nth term
$a_n= (1^2 + 2^2 + 3^2 + ... + n^2) \\ =\frac{n(n+1)(2n+1)}{6}+\frac{n(2n^2+3n+1)}{}+\frac{2n^3+3n^2+n}{6}\\$
Thus $a_n=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n$
Thus,
$S_n=\displaystyle\sum_{k=1}^{n} a_k=\displaystyle\sum_{k=1}^{n} (\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n)\\ =\displaystyle\sum_{k=1}^{n} (\frac{1}{3}n^3)+\displaystyle\sum_{k=1}^{n}\frac{1}{2}n^2+\displaystyle\sum_{k=1}^{n} \frac{1}{6}n\\ =\frac{1}{3}\displaystyle\sum_{k=1}^{n} (n^3)+\displaystyle\sum_{k=1}^{n}\frac{1}{2}n^2+\frac{1}{6}\\ =\frac{1}{3}[\frac{n(n+1)}{2}]^2+\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}]+\frac{1}{6}[\frac{n(n+1)}{2}]\\ =\frac{n(n+1)}{6}[\frac{n(n+1)}{2}+\frac{2n+1}{2}+\frac{1}{2}]\\ =\frac{n(n+1)}{6}(\frac{n^2+n+2n+1+1}{2})\\ =\frac{n(n+1)}{6}(\frac{n^2+n+2n+2}{2})\\ =\frac{n(n+1)}{6}[\frac{(n+1)(n+2)}{2}]\\ =\frac{n(n+1)^2(n+2)}{12}$

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Answer: Please refer to the attached file.

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