Question
For any odd integer $n \geq 1$, $n^{3}-(n-1)^{3}+\ldots+(-1)^{n-1} 1^{3}=$(A) $\frac{1}{2}(n-1)^{2}(2 n-1)$(B) $\frac{1}{4}(n-1)^{2}(2 n-1)$(C) $\frac{1}{2}(n+1)^{2}(2 n-1)$(D) $\frac{1}{4}(n+1)^{2}(2 n-1)$
Step 1
Step 1: We can rewrite the given expression as a sum of cubes: \[n^{3}-(n-1)^{3}+(n-2)^{3}-\ldots+(-1)^{n-1} 1^{3}\] Show more…
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