For any odd integer n ge 1,{n^3} - {(n - 1)^3 | vedclass

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(D) Given the series $S = {n^3} - {(n - 1)^3} + {(n - 2)^3} - {(n - 3)^3} + \dots + {1^3}$.Since $n$ is odd,the terms with even indices are subtracted

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$\frac{1}{4}{(n + 1)^2}(2n - 1)$

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(D) Given the series $S = {n^3} - {(n - 1)^3} + {(n - 2)^3} - {(n - 3)^3} + \dots + {1^3}$.Since $n$ is odd,the terms with even indices are subtracted and terms with odd indices are added.We can rewrite the sum as the sum of all cubes from $1^3$ to $n^3$ minus twice the sum of the cubes of the even terms:$S = \sum_{k=1}^{n} k^3 - 2 \sum_{j=1}^{(n-1)/2} (2j)^3$.Using the formula $\sum_{k=1}^{n} k^3 = [\frac{n(n+1)}{2}]^2$,we get:$S = [\frac{n(n+1)}{2}]^2 - 2 \cdot 8 \sum_{j=1}^{(n-1)/2} j^3$.$S = \frac{n^2(n+1)^2}{4} - 16 [\frac{(\frac{n-1}{2})(\frac{n-1}{2} + 1)}{2}]^2$.$S = \frac{n^2(n+1)^2}{4} - 16 [\frac{(\frac{n-1}{2})(\frac{n+1}{2})}{2}]^2$.$S = \frac{n^2(n+1)^2}{4} - 16 [\frac{(n-1)(n+1)}{8}]^2$.$S = \frac{n^2(n+1)^2}{4} - 16 \cdot \frac{(n-1)^2(n+1)^2}{64}$.$S = \frac{n^2(n+1)^2}{4} - \frac{(n-1)^2(n+1)^2}{4}$.$S = \frac{(n+1)^2}{4} [n^2 - (n-1)^2]$.$S = \frac{(n+1)^2}{4} [n^2 - (n^2 - 2n + 1)]$.$S = \frac{(n+1)^2}{4} (2n - 1)$.

For any odd integer $n \ge 1$,${n^3} - {(n - 1)^3} + ........... + {( - 1)^{n - 1}}{1^3} = $

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