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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

In how many different ways can the letters of the word '$CYCLE$' be arranged?

If $^8C_r = ^8C_{r+2}$,then the value of $^rC_2$ is

The number of four-letter words that can be formed from the letters of the word '$MAYANK$' such that both '$A$'s are included but never together is equal to:

The number of ways in which $5$ male and $2$ female members of a committee can be seated around a round table so that the two female are not seated together is

From a group of $7$ men and $6$ women,five persons are to be selected to form a committee so that at least $3$ men are there on the committee. In how many ways can it be done?

Vedclass Test Series

Exam Paper Generator

Online Exam Module