MHT CET · Maths · Sequences and Series
\(\frac{1^{2}}{2}+\frac{1^{2}+2^{2}}{3}+\frac{1^{2}+2^{2}+3^{2}}{4}+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{5}+\ldots \ldots \ldots \ldots\) upto 8 terms \(=\)
- A 76
- B 74
- C 78
- D 72
Answer & Solution
Correct Answer
(B) 74
Step-by-step Solution
Detailed explanation
\(\frac{1^{2}}{2}+\frac{1^{2}+2^{2}}{3}+\frac{1^{2}+2^{2}+3^{2}}{4}+\)
\(\begin{array}{l}\text { general } \\ \text { term }\end{array} \sum_{t=1}^{8} \frac{\sum \varepsilon^{2}}{\varepsilon}=\)
\(=\sum_{\varepsilon=1}^{8} \frac{\varepsilon(\varepsilon+1)(2 q+1)}{6 \varepsilon}\)
\(=\frac{1}{6} \sum_{\varepsilon=1}^{8} \frac{(2+1)(2 \varepsilon+1)}{(8)}\)
\(=\frac{1}{6} \sum_{\varepsilon=1}^{8}\left[2 \varepsilon^{2}+3 r+1\right]\)
\(=\frac{1}{6}\left[\frac{2 \times 8 \times 9 \times 17}{6}+\frac{3 \times 8 \times 9}{2}+8\right]\)
\(=74\)
\(\begin{array}{l}\text { general } \\ \text { term }\end{array} \sum_{t=1}^{8} \frac{\sum \varepsilon^{2}}{\varepsilon}=\)
\(=\sum_{\varepsilon=1}^{8} \frac{\varepsilon(\varepsilon+1)(2 q+1)}{6 \varepsilon}\)
\(=\frac{1}{6} \sum_{\varepsilon=1}^{8} \frac{(2+1)(2 \varepsilon+1)}{(8)}\)
\(=\frac{1}{6} \sum_{\varepsilon=1}^{8}\left[2 \varepsilon^{2}+3 r+1\right]\)
\(=\frac{1}{6}\left[\frac{2 \times 8 \times 9 \times 17}{6}+\frac{3 \times 8 \times 9}{2}+8\right]\)
\(=74\)
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