sum_{r=1}^{15} r^2 left( frac{{}^{15}C_r}{{}^ | vedclass

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(B) We know that the property of binomial coefficients is $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$.Substituting $n=15$,we get $\frac{{}^{15

Vedclass · Accepted Answer

$680$

Vedclass · Answer

(B) We know that the property of binomial coefficients is $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$.Substituting $n=15$,we get $\frac{{}^{15}C_r}{{}^{15}C_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r}$.Now,the given sum is $S = \sum_{r=1}^{15} r^2 \left( \frac{16-r}{r} ight)$.$S = \sum_{r=1}^{15} r(16-r) = \sum_{r=1}^{15} (16r - r^2)$.$S = 16 \sum_{r=1}^{15} r - \sum_{r=1}^{15} r^2$.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 $n=15$:$\sum_{r=1}^{15} r = \frac{15 	imes 16}{2} = 15 	imes 8 = 120$.$\sum_{r=1}^{15} r^2 = \frac{15 	imes 16 	imes 31}{6} = 5 	imes 8 	imes 31 = 1240$.$S = 16(120) - 1240 = 1920 - 1240 = 680$.

$\sum_{r=1}^{15} r^2 \left( \frac{{}^{15}C_r}{{}^{15}C_{r-1}} \right) = $

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