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JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Binomial Theorem

The value of

\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots . + \frac{^{100} C _{100}}{101}

is :

(A) $\frac{2 ^{101}}{100}$
(B) $\frac{2 ^{100}}{100}$
(C) $\frac{2 ^{101}}{101}$
(D) $\frac{2 ^{100}}{101}$ ✓ Correct

Answer: (D)

Solution

The sum is

S = \sum_{r = 50}^{100} \frac{^{100} C _{r}}{r + 1}

.

Using the identity

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

:

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

.

Since

\sum_{k = 0}^{101}^{101} C_{k} = 2^{101}

and by symmetry

\sum_{k = 51}^{101}^{101} C_{k} = \frac{2 ^{101}}{2} = 2^{100}

,

we get

S = \frac{2 ^{100}}{101}

.

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option

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