MCQ
$\frac{{{1^3} + {2^3} + {3^3} + {4^3} + {{........12}^3}}}{{{1^2} + {2^2} + {3^2} + {4^2} + {{.........12}^2}}} = $
- ✓$\frac{{234}}{{25}}$
- B$\frac{{243}}{{35}}$
- C$\frac{{263}}{{27}}$
- DNone of these
$ = \frac{{\left( {\sum\limits_{n = 1}^{12} {{n^3}} } \right)}}{{\left( {\sum\limits_{n = 1}^{12} {{n^2}} } \right)}} = {\left[ {\frac{{n(n + 1)}}{2}} \right]^2} \times \frac{6}{{n\,(n + 1)\,(2n + 1)}}$
$ = \frac{3}{2}.\frac{{n\,(n + 1)}}{{(2n + 1)}} = \frac{3}{2}.\frac{{12\,\,.\,\,13}}{{25}} = \frac{{234}}{{25}}$, [Putting $n = 12$].
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