For an odd integer n ge 1,the value of n^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 an odd integer,we can group the terms.$S = \sum_{k=1}^{n} (-1)

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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 an odd integer,we can group the terms.$S = \sum_{k=1}^{n} (-1)^{n-k} k^3$.For $n=1$,$S = 1^3 = 1$. Checking options: $D$ gives $\frac{1}{4}(1+1)^2(2(1)-1) = \frac{1}{4}(4)(1) = 1$.For $n=3$,$S = 3^3 - 2^3 + 1^3 = 27 - 8 + 1 = 20$. Checking options: $D$ gives $\frac{1}{4}(3+1)^2(2(3)-1) = \frac{1}{4}(16)(5) = 20$.Thus,the general formula is $\frac{1}{4}(n+1)^2(2n-1)$.

For an odd integer $n \ge 1$,the value of $n^3 - (n-1)^3 + (n-2)^3 - (n-3)^3 + \dots + (-1)^{n-1} 1^3$ is:

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