Sequence and Series – Series Summation – JEE Main 24 Jan 2026 Shift 1

Solution:

The given series is:
$$S = \frac{1}{25!0!} + \frac{1}{23!3!} + \frac{1}{21!5!} + \dots + \frac{1}{1!25!}$$

Multiply and divide by $26!$:
$$S = \frac{1}{26!} \left[ \frac{26!}{1!25!} + \frac{26!}{3!23!} + \frac{26!}{5!21!} + \dots + \frac{26!}{25!1!} \right]$$
$$S = \frac{1}{26!} \left[ ^{26}C_1 + ^{26}C_3 + ^{26}C_5 + \dots + ^{26}C_{25} \right]$$

The expression in the bracket is the sum of binomial coefficients with odd indices.
We know that $^{n}C_1 + ^{n}C_3 + ^{n}C_5 + \dots = 2^{n-1}$.
Here $n=26$.
So, Sum $= 2^{26-1} = 2^{25}$.

Thus,
$$S = \frac{1}{26!} \times 2^{25}$$

Given $13S = \frac{2^k}{n!}$:
$$13 \times \frac{2^{25}}{26!} = \frac{2^k}{n!}$$
$$13 \times \frac{2^{25}}{26 \times 25!} = \frac{2^k}{n!}$$
$$\frac{13}{26} \times \frac{2^{25}}{25!} = \frac{2^k}{n!}$$
$$\frac{1}{2} \times \frac{2^{25}}{25!} = \frac{2^k}{n!}$$
$$\frac{2^{24}}{25!} = \frac{2^k}{n!}$$

Comparing LHS and RHS:
$k = 24$ and $n = 25$.

We need $n+k$:
$$n+k = 25 + 24 = 49$$

Ans. (3)