Question

For any odd integer $n⩾1,$ prove that

$n^3-{\left(n-1\right)}^3+......+{\left(-1\right)}^{n-1}{1}^3=\frac{1}{4}{(n+1)}^2(2n-1)$

Answer 1

$\because n$ is an odd integer ${(-1)}^{n-1}=1$ for $n=1$

$n-3,n-5,n-6,...$

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

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

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

$\because n-1,n-3$ are even integers.

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

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

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