Binomial Theorem 1 Question 28
30. For any odd integer $n \geq 1, n^{3}-(n-1)^{3}+\ldots$
$+(-1)^{n-1} 1^{3}=\ldots$
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Solution:
- Since, $n$ is an odd integer, $(-1)^{n-1}=1$
and $n-1, n-3, n-5$, etc., are even integers, then
$$ \begin{aligned} n^{3}- & (n-1)^{3}+(n-2)^{3}-(n-3)^{3}+\ldots+(-1)^{n-1} \cdot 1^{3} \\ & =n^{3}+(n-1)^{3}+(n-2)^{3}+\ldots+1^{3} \\ & \left.-2\left[(n-1)^{3}+(n-3)^{3}+\ldots+2^{3}\right)\right] \\ & =\Sigma n^{3}-2 \times 2^{3} \frac{n-1}{2}+\frac{n-3}{2}^{3}+\ldots+1^{3} \end{aligned} $$
$[\because n-1, n-3, \ldots$, are even integers $]$
$$ =\Sigma n^{3}-16 \quad \Sigma \frac{n-1}{2}^{3} $$
$=\frac{n(n+1)}{2}^{2}-16 \frac{1}{2} \frac{n-1}{2} \quad \frac{n-1}{2}+1^{2}$
$=\frac{1}{4} n^{2}(n+1)^{2}-\frac{16(n-1)^{2}(n+1)^{2}}{4 \times 4 \times 4}$
$=\frac{1}{4}(n+1)^{2}\left[n^{2}-(n-1)^{2}\right]=\frac{1}{4}(n+1)^{2}(2 n-1)$
