JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

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

Solution

Each term has the form $\frac{1}{( 2 k - 1 )! ( 27 - 2 k )!}$ with denominator summing to $26$ . Multiply and divide by $26!$ :

$S = \frac{1}{26 !} [(1 26) + (3 26) + (5 26) + \dots + (25 26)]$ .

This is the sum of binomial coefficients with odd lower index from $1$ to $25$ , exactly $13$ terms, equal to $2^{25}$ (half of $2^{26}$ ).

$\Rightarrow S = \frac{2 ^{25}}{26 !}$ .

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

Hence $k = 24$ , $n = 25$ , $n + k = 49$ .

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