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The value of $\sum\limits_{r = 1}^{15} {{r^2}\,\left( {\frac{{^{15}{C_r}}}{{^{15}{C_{r - 1}}}}} \right)} $ is equal to
$1240$
$560$
$1085$
$680$
Solution
$\sum\limits_{r = 1}^{15} {{r^2}} \,\left( {\frac{{{{15}_{{C_r}}}}}{{{{15}_{{C_{r – 1}}}}}}} \right)$
$\frac{{{{15}_{{C_r}}}}}{{{{15}_{C – 1}}}} = \frac{{\frac{{\frac{{15!}}{{15! – rr!}}}}{{15!}}}}{{r – 1!\,16 – r!}}$
$ = \frac{{\frac{{\frac{1}{{15 – r!\,r – 1!}}}}{1}}}{{r – 1!\,15 – r.(16 – r)!}}$
$ = \frac{{16 – r}}{r}$
$ = \sum\limits_{r = 1}^{15} {{r^2}} \left( {\frac{{16 – r}}{r}} \right) = \sum\limits_{r = 1}^{15} {r(16 – r)} $
$ = 16\sum\limits_{r = 1}^{15} {r – } \sum\limits_{r = 1}^{15} {{r^2}} $
$ = \frac{{16 \times 15 \times 16}}{2} – \frac{{15 \times 31 \times 16}}{6}$
$ = 8 \times 15 \times 16 – 5 \times 8 \times 31 = 1920 – 1240 = 680$
