The sum frac{3 times 1}{1^2} + frac{5 times ( | vedclass

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(B) The $n^{th}$ term of the series is given by:$T_n = \frac{(2n + 1) \sum_{k=1}^n k^3}{\sum_{k=1}^n k^2}$Using the formulas $\sum k^3 = \frac{n^2(n+1

Vedclass · Accepted Answer

$660$

Vedclass · Answer

(B) The $n^{th}$ term of the series is given by:$T_n = \frac{(2n + 1) \sum_{k=1}^n k^3}{\sum_{k=1}^n k^2}$Using the formulas $\sum k^3 = \frac{n^2(n+1)^2}{4}$ and $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$T_n = \frac{(2n + 1) \cdot \frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(2n+1)}{6}}$$T_n = \frac{n^2(n+1)^2}{4} \cdot \frac{6}{n(n+1)} = \frac{3}{2} n(n+1)$Now,the sum of the first $10$ terms is:$S_{10} = \sum_{n=1}^{10} \frac{3}{2} (n^2 + n) = \frac{3}{2} \left[ \sum_{n=1}^{10} n^2 + \sum_{n=1}^{10} n ight]$$S_{10} = \frac{3}{2} \left[ \frac{10(11)(21)}{6} + \frac{10(11)}{2} ight]$$S_{10} = \frac{3}{2} [385 + 55] = \frac{3}{2} [440] = 3 	imes 220 = 660$.

The sum $\frac{3 \times 1}{1^2} + \frac{5 \times (1^3 + 2^3)}{1^2 + 2^2} + \frac{7 \times (1^3 + 2^3 + 3^3)}{1^2 + 2^2 + 3^2} + \dots$ up to the $10^{th}$ term is:

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