The value of sumlimits_{r = 1}^{15} {{r^2},le | vedclass

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(D) We know that $\frac{^{n}C_{r}}{^{n}C_{r-1}} = \frac{n-r+1}{r}$.For $n = 15$,we have $\frac{^{15}C_{r}}{^{15}C_{r-1}} = \frac{15-r+1}{r} = \frac{16

Vedclass · Accepted Answer

$680$

Vedclass · Answer

(D) We know that $\frac{^{n}C_{r}}{^{n}C_{r-1}} = \frac{n-r+1}{r}$.For $n = 15$,we have $\frac{^{15}C_{r}}{^{15}C_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r}$.Substituting this into the summation:$\sum\limits_{r = 1}^{15} r^2 \left( \frac{16-r}{r} ight) = \sum\limits_{r = 1}^{15} r(16-r) = \sum\limits_{r = 1}^{15} (16r - r^2)$.This can be split into two sums:$16 \sum\limits_{r = 1}^{15} r - \sum\limits_{r = 1}^{15} r^2$.Using the formulas $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=15$:$16 \left( \frac{15 	imes 16}{2} ight) - \left( \frac{15 	imes 16 	imes 31}{6} ight)$.$= 16(120) - (5 	imes 8 	imes 31) = 1920 - 1240 = 680$.

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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