Binomial TheoremHard

Question

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

Options

A.$\frac{2^{101}}{100}$
B.$\frac{2^{100}}{100}$
C.$\frac{2^{101}}{101}$
D.$\frac{2^{100}}{101}$

Solution

$S = \sum_{r = 50}^{100}\mspace{2mu}\frac{\ ^{100}C_{r}}{r + 1} = \sum_{r = 50}^{100}\mspace{2mu}\frac{1}{r + 1} \cdot \frac{r + 1}{101} \cdot \ ^{101}C_{r + 1}$

$$\begin{matrix} & S = \frac{1}{101}\sum_{r = 50}^{100}\mspace{2mu}\mspace{2mu}\ ^{101}C_{r + 1} \\ & \ = \frac{1}{101} \times \frac{2^{101}}{2} = \frac{2^{100}}{101} \end{matrix}$$

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