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$\frac{1^{2}}{2}+\frac{1^{2}+2^{2}}{3}+\frac{1^{2}+2^{2}+3^{2}}{4}+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{5}+\ldots \ldots \ldots \ldots$ upto 8 terms $=$
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74
$\frac{1^{2}}{2}+\frac{1^{2}+2^{2}}{3}+\frac{1^{2}+2^{2}+3^{2}}{4}+$
$\begin{array}{l}\text { general } \\ \text { term }\end{array} \sum_{t=1}^{8} \frac{\sum \varepsilon^{2}}{\varepsilon}=$
$=\sum_{\varepsilon=1}^{8} \frac{\varepsilon(\varepsilon+1)(2 q+1)}{6 \varepsilon}$
$=\frac{1}{6} \sum_{\varepsilon=1}^{8} \frac{(2+1)(2 \varepsilon+1)}{(8)}$
$=\frac{1}{6} \sum_{\varepsilon=1}^{8}\left[2 \varepsilon^{2}+3 r+1\right]$
$=\frac{1}{6}\left[\frac{2 \times 8 \times 9 \times 17}{6}+\frac{3 \times 8 \times 9}{2}+8\right]$
$=74$
$\begin{array}{l}\text { general } \\ \text { term }\end{array} \sum_{t=1}^{8} \frac{\sum \varepsilon^{2}}{\varepsilon}=$
$=\sum_{\varepsilon=1}^{8} \frac{\varepsilon(\varepsilon+1)(2 q+1)}{6 \varepsilon}$
$=\frac{1}{6} \sum_{\varepsilon=1}^{8} \frac{(2+1)(2 \varepsilon+1)}{(8)}$
$=\frac{1}{6} \sum_{\varepsilon=1}^{8}\left[2 \varepsilon^{2}+3 r+1\right]$
$=\frac{1}{6}\left[\frac{2 \times 8 \times 9 \times 17}{6}+\frac{3 \times 8 \times 9}{2}+8\right]$
$=74$
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