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Find the sum to $n$ terms of the series $1^2+(1^2+2^2)+(1^2+2^2+3^2)+........$

$\begin{array}{1 1}\large\frac{n(n+1)(n+2)^2}{12}\\\large\frac{n^2(n+1)(n+2)}{12} \\\large\frac{n(n+1)^2(n+2)}{12} \\ \large\frac{n(n+1)^2(n+2)}{24}\end{array} $

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  • $1^2+2^2+3^2+.......=\sum \:n^2=\large\frac{n(n+1)(2n+1)}{6}$
  • Sum of $n$ terms of any series $S_n=\sum t_n$
  • $\sum n^3=\large\frac{n^2(n+1)^2}{4}$
  • $\sum n=\large\frac{n(n+1)}{2}$
Given series is $1^2+(1^2+2^2)+(1^2+2^2+3^2)+........$
We have to find the $n^{th}$ term $t_n$ of the series
Clearly $n^{th}$ term of the series is
$1^2+2^2+3^2+............n^2=\sum n^2$
We know that $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
$\Rightarrow\:t_n=\sum n^2=\large\frac{n(n+1)(2n+1)}{6}=\large\frac{1}{6}$$(2n^3+3n^2+n)$
We know that sum of $n$ terms of any series $S_n=\sum t_n$
$\Rightarrow\:S_n=\sum t_n=\sum \large\frac{1}{6}$$(2n^3+3n^2+n)$
$\qquad=\large\frac{1}{6}$$\big[\sum2n^3+\sum 3n^2+\sum n\big]$
$\qquad=\large\frac{1}{6}$$\bigg[2\times\large\frac{n^2(n+1)^2}{4}$$+3\times \large\frac{n(n+1)(2n+1)}{6}$$+\large\frac{n(n+1)}{2}\bigg]$
answered Mar 24, 2014 by rvidyagovindarajan_1

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