The sum of the series frac{2^{3}-1^{3}}{1 tim | vedclass

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Question

(C) Let the $n^{th}$ term be $T_n$. The numerator of the $n^{th}$ term is $S_n = \sum_{k=1}^{2n} (-1)^{k-1} k^3 = 1^3 - 2^3 + 3^3 - 4^3 + \ldots + (2n

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) Let the $n^{th}$ term be $T_n$. The numerator of the $n^{th}$ term is $S_n = \sum_{k=1}^{2n} (-1)^{k-1} k^3 = 1^3 - 2^3 + 3^3 - 4^3 + \ldots + (2n)^3$. This can be written as $S_n = \sum_{k=1}^{n} ((2k)^3 - (2k-1)^3)$. Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$,we have $(2k)^3 - (2k-1)^3 = (1)(4k^2 + 2k(2k-1) + (2k-1)^2) = 4k^2 + 4k^2 - 2k + 4k^2 - 4k + 1 = 12k^2 - 6k + 1$. Thus,$S_n = \sum_{k=1}^{n} (12k^2 - 6k + 1) = 12 \frac{n(n+1)(2n+1)}{6} - 6 \frac{n(n+1)}{2} + n = 2n(n+1)(2n+1) - 3n(n+1) + n = n(4n^2 + 6n + 2 - 3n - 3 + 1) = n(4n^2 + 3n) = n^2(4n+3)$. The denominator of the $n^{th}$ term is $n(4n+3)$. Therefore,$T_n = \frac{n^2(4n+3)}{n(4n+3)} = n$. The sum of the series is $\sum_{n=1}^{15} T_n = \sum_{n=1}^{15} n = \frac{15 	imes 16}{2} = 120$.

The sum of the series $\frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\ldots + \frac{30^{3}-29^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$ is equal to:

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