frac{1^{2}}{2} + frac{1^{2}+2^{2}}{3} + frac{ | vedclass

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(B) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^{n} k^2}{n+1}$.Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Vedclass · Accepted Answer

$74$

Vedclass · Answer

(B) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^{n} k^2}{n+1}$.Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$T_n = \frac{n(n+1)(2n+1)}{6(n+1)} = \frac{n(2n+1)}{6} = \frac{2n^2 + n}{6}$.We need to find the sum of the first $8$ terms,$S_8 = \sum_{n=1}^{8} T_n = \sum_{n=1}^{8} \frac{2n^2 + n}{6}$.$S_8 = \frac{1}{6} \left[ 2 \sum_{n=1}^{8} n^2 + \sum_{n=1}^{8} n \right]$.Using $\sum_{n=1}^{8} n^2 = \frac{8(9)(17)}{6} = 204$ and $\sum_{n=1}^{8} n = \frac{8(9)}{2} = 36$.$S_8 = \frac{1}{6} [2(204) + 36] = \frac{1}{6} [408 + 36] = \frac{444}{6} = 74$.

$\frac{1^{2}}{2} + \frac{1^{2}+2^{2}}{3} + \frac{1^{2}+2^{2}+3^{2}}{4} + \frac{1^{2}+2^{2}+3^{2}+4^{2}}{5} + \dots$ up to $8$ terms $=$

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