Q28 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
Let $\mathrm{S}=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots$ up to 13 terms. If $13 \mathrm{~S}=\frac{2^{k}}{n!}, k \in \mathrm{~N}$, then $n+k$ is equal to
- A50
- Correct answerB49
- C52
- D51
Solution
The general term is $\frac{1}{(2r-1)!(27-2r)!}$ for $r=1$ to $13$.
Multiply and divide by $26!$: $\frac{1}{(2r-1)!(27-2r)!} = \frac{1}{26!}\binom{26}{2r-1}$
$S = \frac{1}{26!}\sum_{r=1}^{13}\binom{26}{2r-1} = \frac{1}{26!}\left[\binom{26}{1}+\binom{26}{3}+\cdots+\binom{26}{25}\right]$
Sum of odd-indexed binomial coefficients: $\sum \binom{26}{\text{odd}} = 2^{25}$
$S = \frac{2^{25}}{26!}$
$13S = \frac{13 \cdot 2^{25}}{26!} = \frac{13 \cdot 2^{25}}{26 \cdot 25!} = \frac{2^{25}}{2 \cdot 25!} = \frac{2^{24}}{25!}$
So $k = 24,\; n = 25$, giving $n + k = 49$.
Multiply and divide by $26!$: $\frac{1}{(2r-1)!(27-2r)!} = \frac{1}{26!}\binom{26}{2r-1}$
$S = \frac{1}{26!}\sum_{r=1}^{13}\binom{26}{2r-1} = \frac{1}{26!}\left[\binom{26}{1}+\binom{26}{3}+\cdots+\binom{26}{25}\right]$
Sum of odd-indexed binomial coefficients: $\sum \binom{26}{\text{odd}} = 2^{25}$
$S = \frac{2^{25}}{26!}$
$13S = \frac{13 \cdot 2^{25}}{26!} = \frac{13 \cdot 2^{25}}{26 \cdot 25!} = \frac{2^{25}}{2 \cdot 25!} = \frac{2^{24}}{25!}$
So $k = 24,\; n = 25$, giving $n + k = 49$.
Hence, the correct answer is option: 'B'.