The sum 1 + frac{1}{2}(1^2+2^2) + frac{1}{3}( | vedclass

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(C) The $n$-th term of the series is given by $T_n = \frac{1}{n} \sum_{k=1}^n k^2$.Using the formula for the sum of squares of the first $n$ natural n

Vedclass · Accepted Answer

$\frac{315}{2}$

Vedclass · Answer

(C) The $n$-th term of the series is given by $T_n = \frac{1}{n} \sum_{k=1}^n k^2$.Using the formula for the sum of squares of the first $n$ natural numbers, $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.Thus, $T_n = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6} = \frac{2n^2+3n+1}{6} = \frac{n^2}{3} + \frac{n}{2} + \frac{1}{6}$.To find the sum of the first $10$ terms, we calculate $S_{10} = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} (\frac{n^2}{3} + \frac{n}{2} + \frac{1}{6})$.$S_{10} = \frac{1}{3} \sum_{n=1}^{10} n^2 + \frac{1}{2} \sum_{n=1}^{10} n + \sum_{n=1}^{10} \frac{1}{6}$.$S_{10} = \frac{1}{3} \left( \frac{10 \cdot 11 \cdot 21}{6} \right) + \frac{1}{2} \left( \frac{10 \cdot 11}{2} \right) + \frac{10}{6}$.$S_{10} = \frac{1}{3} (385) + \frac{55}{2} + \frac{5}{3} = \frac{385+5}{3} + 27.5 = \frac{390}{3} + 27.5 = 130 + 27.5 = 157.5$.Since $157.5 = \frac{315}{2}$, the correct option is $C$.

The sum $1 + \frac{1}{2}(1^2+2^2) + \frac{1}{3}(1^2+2^2+3^2) + \dots$ up to $10$ terms is equal to:

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