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

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(D) Let $S = {n^3} - {(n - 1)^3} + {(n - 2)^3} - {(n - 3)^3} + \dots + {1^3}$.Since $n$ is odd,the last term is ${1^3}$.We can write $S = \sum_{k=1}^{

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

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(D) Let $S = {n^3} - {(n - 1)^3} + {(n - 2)^3} - {(n - 3)^3} + \dots + {1^3}$.Since $n$ is odd,the last term is ${1^3}$.We can write $S = \sum_{k=1}^{n} k^3 - 2 \sum_{k=1}^{(n-1)/2} (2k)^3$.Using the formula $\sum_{k=1}^{m} k^3 = \left[ \frac{m(m+1)}{2} ight]^2$,we get:$S = \left[ \frac{n(n+1)}{2} ight]^2 - 2 	imes 8 \sum_{k=1}^{(n-1)/2} k^3$$S = \frac{n^2(n+1)^2}{4} - 16 \left[ \frac{\frac{n-1}{2} (\frac{n-1}{2} + 1)}{2} ight]^2$$S = \frac{n^2(n+1)^2}{4} - 16 \left[ \frac{(n-1)(n+1)}{8} ight]^2$$S = \frac{n^2(n+1)^2}{4} - 16 \frac{(n-1)^2(n+1)^2}{64}$$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} + \dots + {( - 1)^{n - 1}}{1^3} = $

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