MCQ
For any odd integer $n \ge 1$, ${n^3} - {(n - 1)^3} + ........... + {( - 1)^{n - 1}}{1^3} = $
- A$\frac{1}{2}{(n - 1)^2}(2n - 1)$
- B$\frac{1}{4}{(n - 1)^2}(2n - 1)$
- C$\frac{1}{2}{(n + 1)^2}(2n - 1)$
- ✓$\frac{1}{4}{(n + 1)^2}(2n - 1)$
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]$
[$\;n - 1,\;n - 3$ 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}] $
$= \frac{1}{4}{(n + 1)^2}(2n - 1)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.