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JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Binomial Theorem

Let

S = \frac{1}{25 !} + \frac{1}{3 ! 23 !} + \frac{1}{5 ! 21 !} + \dots

up to 13 terms.

If

13 S = \frac{2 ^{k}}{n !}, k \in N

, then

n + k

is equal to

(A)51
(B)52
(C)49✓ Correct
(D)50

Answer: (C)

Solution

Multiplying by

26!

,

S = \frac{1}{26 !} [(1 26) + (3 26) + \dots + (25 26)]

, the sum of the

13

odd-order binomial coefficients, which equals

2^{25}

. Hence

S = \frac{2 ^{25}}{26 !}

and

13 S = \frac{13 \cdot 2 ^{25}}{26 \cdot 25 !} = \frac{2 ^{24}}{25 !}

. Comparing with

\frac{2 ^{k}}{n !}

gives

k = 24, n = 25

, so

n + k = 49

.

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option

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