The value of ^ 100 C_ 50 51 + ^ 100 C_ 51 52 + .+ ^ 100 C_ 100 101 is :

Correct answer: D. 2^ 100 101 Using n r r+1 = (1)/(n+1) n+1 r+1 : ∑_ r=50 ^ 100 100 r r+1 = (1)/(101)∑_ r=50 ^ 100 101 r+1 = (1)/(101)∑_ k=51 ^ 101 101 k . Since 101 is odd, by symmetry ∑_ k=51 ^ 101 101 k = ∑_ k=0 ^ 50 101 k = 2^ 101 2 = 2^ 100 . Required value = 2^ 100 101 .

JEE Main 202623 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Binomial Theorem

The value of ^ 100 C_ 50 51 + ^ 100 C_ 51 52 + .+ ^ 100 C_ 100 101...

The value of

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

is :

Options

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

Correct answer

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

Step-by-step solution

Using

\frac{\binom{n}{r}}{r+1} = \frac{1}{n+1}\binom{n+1}{r+1}

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

.
Since

101

is odd, by symmetry

\sum_{k=51}^{101}\binom{101}{k} = \sum_{k=0}^{50}\binom{101}{k} = \frac{2^{101}}{2} = 2^{100}

.
Required value

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

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