$\frac{1^{2}}{2}+\frac{1^{2}+2^{2}}{3}+\frac{1^{2}+2^{2}+3^{2}}{4}+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{5}+\ldots \ldots \ldots \ldots$ upto 8 terms $=$

Question

Aicharya · Accepted Answer

$\frac{1^{2}}{2}+\frac{1^{2}+2^{2}}{3}+\frac{1^{2}+2^{2}+3^{2}}{4}+$ $\begin{array}{l}	ext { general } \ 	ext { term }\end{array} \sum_{t=1}^{8} \frac{\sum \varepsilon^{2}}{\varepsilon}=$ $=\sum_{\varepsilon=1}^{8} \frac{\varepsilon(\varepsilon+1)(2 q+1)}{6 \varepsilon}$ $=\frac{1}{6} \sum_{\varepsilon=1}^{8} \frac{(2+1)(2 \varepsilon+1)}{(8)}$ $=\frac{1}{6} \sum_{\varepsilon=1}^{8}\left[2 \varepsilon^{2}+3 r+1ight]$ $=\frac{1}{6}\left[\frac{2 	imes 8 	imes 9 	imes 17}{6}+\frac{3 	imes 8 	imes 9}{2}+8ight]$ $=74$

$\frac{1^{2}}{2}+\frac{1^{2}+2^{2}}{3}+\frac{1^{2}+2^{2}+3^{2}}{4}+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{5}+\ldots…

Solution