Q74 / 150 · MHT CET 2020 (14 Oct Shift 2) · Mathematics

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

  • A
    76
  • Correct answer
    B
    74
  • C
    78
  • D
    72

Solution

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

Hence, the correct answer is option: 'B'.

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