The value of sumlimits_{r = 1}^{15} {{r^ | vedclass

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Question

$\sum\limits_{r = 1}^{15} {{r^2}} \,\left( {\frac{{{{15}_{{C_r}}}}}{{{{15}_{{C_{r - 1}}}}}}} \right)$

$\frac{{{{15}_{{C_r}}}}}{{{{15}_{C - 1}}}} =

Vedclass · Accepted Answer

$680$

Vedclass · Answer

$\sum\limits_{r = 1}^{15} {{r^2}} \,\left( {\frac{{{{15}_{{C_r}}}}}{{{{15}_{{C_{r - 1}}}}}}} ight)$

$\frac{{{{15}_{{C_r}}}}}{{{{15}_{C - 1}}}} = \frac{{\frac{{\frac{{15!}}{{15! - rr!}}}}{{15!}}}}{{r - 1!\,16 - r!}}$

$ = \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}} ight) = \sum\limits_{r = 1}^{15} {r(16 - r)} $

$ = 16\sum\limits_{r = 1}^{15} {r - } \sum\limits_{r = 1}^{15} {{r^2}} $

$ = \frac{{16 	imes 15 	imes 16}}{2} - \frac{{15 	imes 31 	imes 16}}{6}$

$ = 8 	imes 15 	imes 16 - 5 	imes 8 	imes 31 = 1920 - 1240 = 680$

The value of $\sum\limits_{r = 1}^{15} {{r^2}\,\left( {\frac{{^{15}{C_r}}}{{^{15}{C_{r - 1}}}}} \right)} $ is equal to

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