Question

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

Answer 1

Since $n$ is an odd integer, $(-1{)}^{n-1}=1$ and $n-1,n-3,n-5,....$ are even integers. The given series is

$n^3-(n-1{)}^3+(n-2{)}^3-(n-3{)}^3+...+(-1{)}^{n-1}{1}^3$

$=[n^3+(n-1{)}^3+(n-2{)}^3+...+1^3]-2\left[\right(n-1{)}^3+(n-3{)}^3+...+2^3]$

$=[\frac{n(n+1)}{2}{]}^2-16[\{\frac{1}{2}(\frac{n-1}{2})(\frac{n-1}{2}+1)\}{]}^2$

$=\frac{1}{4}{n}^2(n+1{)}^2-16\frac{(n-1{)}^2(n+1{)}^2}{16\times 4}$

$=\frac{1}{4}(n-1{)}^2[n^2-(n-1{)}^2]$

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