• # question_answer For any odd integer $n\ge 1$,${{n}^{3}}-{{(n-1)}^{3}}+...........+{{(-1)}^{n-1}}{{1}^{3}}=$ [IIT 1996] A) $\frac{1}{2}{{(n-1)}^{2}}(2n-1)$ B) $\frac{1}{4}{{(n-1)}^{2}}(2n-1)$ C) $\frac{1}{2}{{(n+1)}^{2}}(2n-1)$ D) $\frac{1}{4}{{(n+1)}^{2}}(2n-1)$

Since $n$ is an odd integer ${{(-1)}^{n-1}}=1$ and $n-1$, $n-3,\ n-5$ etc. are even integers. We have = ${{n}^{3}}-{{(n-1)}^{3}}+{{(n-2)}^{3}}-{{(n-3)}^{3}}$+$........+{{(-1)}^{n-1}}{{1}^{3}}$ $={{n}^{3}}+{{(n-1)}^{3}}$$+{{(n-2)}^{3}}+.......+{{1}^{3}}$                                    $-2[{{(n-1)}^{3}}+{{(n-3)}^{3}}+.......+{{2}^{3}}]$ $={{n}^{3}}+{{(n-1)}^{3}}+{{(n-2)}^{3}}+.......+{{1}^{3}}$                    $-2\times {{2}^{3}}\left[ {{\left( \frac{n-1}{2} \right)}^{3}}+{{\left( \frac{n-3}{2} \right)}^{3}}+.......+{{1}^{3}} \right]$                                    [$\because \ n-1,\ n-3$are even integers] $={{\left[ \frac{n\,(n+1)}{2} \right]}^{2}}-16{{\left[ \frac{1}{2}\left( \frac{n-1}{2} \right)\,\left( \frac{n-1}{2}+1 \right) \right]}^{2}}$ $=\frac{1}{4}{{n}^{2}}{{(n+1)}^{2}}-16\frac{{{(n-1)}^{2}}{{(n+1)}^{2}}}{16\times 4}$ $=\frac{1}{4}{{(n+1)}^{2}}[{{n}^{2}}-{{(n-1)}^{2}}]=\frac{1}{4}{{(n+1)}^{2}}(2n-1)$.