JEE Main 2026 Jan 23 Shift 1, Mathematics Q12: Properties Of Binomial Coefficient

JEE Main2026Jan 23, Shift 1Mathematics

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

Solution

Use the identity $\frac{^{n} C _{r}}{r + 1} = \frac{1}{n + 1} \cdot^{n + 1} C_{r + 1}$ :

$\frac{^{100} C _{r}}{r + 1} = \frac{1}{101} \cdot^{101} C_{r + 1}$

Thus the sum becomes:

$S = \frac{1}{101} \sum_{r = 50}^{100}^{101} C_{r + 1} = \frac{1}{101} \sum_{k = 51}^{101}^{101} C_{k}$

By symmetry $\sum_{k = 0}^{101}^{101} C_{k} = 2^{101}$ , and the upper half $\sum_{k = 51}^{101} = \frac{2 ^{101}}{2} = 2^{100}$ .

$∴ S = \frac{2 ^{100}}{101}$ .

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