যোগফল  নির্নয় করোঃ

  $1^{2} + \left ( 1^{2} + 2^{2} \right ) + \left ( 1^{2} + 2^{2} + 3^{2} \right ) + \ldots . . u p \to n t e r m s .$  

সমাধান: $n$ তম পদ, $u_{n}=1^{2}+2^{2}+\cdots \cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6}=\frac{1}{6}\left(n+3 n^{2}+2 n^{3}\right)$

$\begin{array}{l} S_{n}=\frac{1}{6} \sum n+\frac{1}{2} \sum n^{2}+\frac{1}{3} \sum n^{3}=\frac{1}{6} \frac{n(n+1)}{2}+\frac{1}{2} \frac{n(n+1)(2 n+1)}{6}+\frac{1}{3} \frac{n^{2}(n+1)^{2}}{4} \\ =\frac{1}{12} n(n+1)\{1+2 n+1+n(n+1)\}=\frac{1}{12}n(n+1)\left(n^{2}+3 n+2\right)=\frac{1}{12} n(n+1)\left(n^{2}+2 n+n+2\right) \\ =\frac{1}{12} n(n+1)(n+1)(n+2)=\frac{1}{12} n(n+1)^{2}(n+2)(\text { Ans. }) \end{array}$