If ^nC_r = C_r and 2 frac{C_1}{C_0} + 4 frac{ | vedclass

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

Question

(B) The general term of the series is $T_r = 2r \frac{C_r}{C_{r-1}}$.Using the property $\frac{C_r}{C_{r-1}} = \frac{n-r+1}{r}$,we get:$T_r = 2r \left

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(B) The general term of the series is $T_r = 2r \frac{C_r}{C_{r-1}}$.Using the property $\frac{C_r}{C_{r-1}} = \frac{n-r+1}{r}$,we get:$T_r = 2r \left( \frac{n-r+1}{r} ight) = 2(n-r+1)$.The sum is $\sum_{r=1}^n T_r = \sum_{r=1}^n 2(n-r+1) = 2 \left[ n^2 - \frac{n(n+1)}{2} + n ight] = 2 \left[ n^2 - \frac{n^2+n}{2} + n ight] = 2 \left[ \frac{2n^2 - n^2 - n + 2n}{2} ight] = n(n+1)$.Given $n(n+1) = 650$,we have $n^2 + n - 650 = 0$.Solving for $n$,$(n+26)(n-25) = 0$,so $n = 25$.Therefore,$^nC_2 = ^{25}C_2 = \frac{25 	imes 24}{2} = 300$.

If $^nC_r = C_r$ and $2 \frac{C_1}{C_0} + 4 \frac{C_2}{C_1} + 6 \frac{C_3}{C_2} + \dots + 2n \frac{C_n}{C_{n-1}} = 650$,then $^nC_2 =$

Similar Questions

If ${}^{21}C_1 + 3 \cdot {}^{21}C_3 + 5 \cdot {}^{21}C_5 + \dots + 19 \cdot {}^{21}C_{19} + 21 \cdot {}^{21}C_{21} = k$,then the number of prime factors of $k$ is:

$\sum_{r=0}^{10} {}^{40-r} C_5$ is equal to

For $z \in \mathbb{C}$,if $(1+z)^n = 1 + { }^n C_1 z + { }^n C_2 z^2 + \ldots + { }^n C_n z^n$ and $\sum_{r=0}^{100} { }^{100} C_r \sin(rx) = \left(2 \cos \frac{x}{2}\right)^{100} \sin(kx)$,then $k =$

If $P_{n}$ denotes the product of the binomial coefficients in the expansion of $(1+x)^{n}$,then $\frac{P_{n+1}}{P_n}=$

The sum of the series $1 + \frac{1}{2} {}^{n}C_{1} + \frac{1}{3} {}^{n}C_{2} + \dots + \frac{1}{n+1} {}^{n}C_{n}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module