The value of frac{^{100}C_{50}}{51} + frac{^{ | vedclass

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

Question

(D) We use the identity $\frac{^{n}C_{r}}{r+1} = \frac{^{n+1}C_{r+1}}{n+1}$.
Given the sum $S = \sum_{r=50}^{100} \frac{^{100}C_{r}}{r+1}$.
Using th

Vedclass · Accepted Answer

$\frac{2^{100}}{101}$

Vedclass · Answer

(D) We use the identity $\frac{^{n}C_{r}}{r+1} = \frac{^{n+1}C_{r+1}}{n+1}$.
Given the sum $S = \sum_{r=50}^{100} \frac{^{100}C_{r}}{r+1}$.
Using the identity, we get $S = \sum_{r=50}^{100} \frac{^{101}C_{r+1}}{101}$.
$S = \frac{1}{101} \sum_{k=51}^{101} {^{101}C_{k}}$, where $k = r+1$.
The sum $\sum_{k=51}^{101} {^{101}C_{k}}$ represents the sum of the last half of the binomial coefficients of $(1+x)^{101}$.
Since $\sum_{k=0}^{101} {^{101}C_{k}} = 2^{101}$ and $\sum_{k=0}^{50} {^{101}C_{k}} = \sum_{k=51}^{101} {^{101}C_{k}} = \frac{2^{101}}{2} = 2^{100}$.
Thus, $S = \frac{2^{100}}{101}$.

The value of $\frac{^{100}C_{50}}{51} + \frac{^{100}C_{51}}{52} + \dots + \frac{^{100}C_{100}}{101}$ is:

Similar Questions

The ratio of the coefficients of the terms $x^{n-r}a^r$ and $x^ra^{n-r}$ in the binomial expansion of $(x+a)^n$ is:

If $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \ldots + (101) \cdot C_{25} = 2^{25} \cdot k$,then $k$ is equal to:

If $C_j = {}^{n}C_j$,then $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = $

If $p$ and $q$ are positive integers,then the coefficients of $x^p$ and $x^q$ in the expansion of $(1 + x)^{p + q}$ are

The value of $\sum_{r=0}^{6} \left({}^{6}C_{r} \cdot {}^{6}C_{6-r}\right)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module