The sum of the following series 1 + 6 + frac{ | vedclass

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(A) The $n^{th}$ term of the series is given by $T_n = \frac{(3 + (n-1) 	imes 3)(1^2 + 2^2 + ... + n^2)}{2n + 1}$.Simplifying $T_n$:$T_n = \frac{3n \

Vedclass · Accepted Answer

$7820$

Vedclass · Answer

(A) The $n^{th}$ term of the series is given by $T_n = \frac{(3 + (n-1) 	imes 3)(1^2 + 2^2 + ... + n^2)}{2n + 1}$.Simplifying $T_n$:$T_n = \frac{3n 	imes \frac{n(n+1)(2n+1)}{6}}{2n+1} = \frac{n^2(n+1)}{2} = \frac{n^3 + n^2}{2}$.The sum of $15$ terms is $S_{15} = \sum_{n=1}^{15} T_n = \frac{1}{2} \sum_{n=1}^{15} (n^3 + n^2)$.Using the summation formulas $\sum n^3 = [\frac{n(n+1)}{2}]^2$ and $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$:$S_{15} = \frac{1}{2} [(\frac{15 	imes 16}{2})^2 + \frac{15 	imes 16 	imes 31}{6}]$.$S_{15} = \frac{1}{2} [120^2 + 1240] = \frac{1}{2} [14400 + 1240] = \frac{15640}{2} = 7820$.

The sum of the following series $1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + ... + 5^2)}{11} + ...$ up to $15$ terms is:

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