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

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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:

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