JEE Mathematics 2026 Question with Solution

Given:

$$S=\frac{1}{2!5!}+\frac{1}{3!2!3!}+\frac{1}{5!2!1!}+\cdots$$

up to $13$ terms, and

$$13S=\frac{2^k}{n!}$$

Find: $n+k$

From the solution pattern, each term is taken as

$$\frac{1}{(r+1)!\,(6-r)!},\quad r=0,1,2,\ldots,12$$

So,

$$S=\sum_{r=0}^{12}\frac{1}{(r+1)!(6-r)!}$$

Now write each term using binomial coefficients:

$$\frac{1}{(r+1)!(6-r)!}=\frac{1}{7!}\binom{7}{r+1}$$

Hence,

$$S=\frac{1}{7!}\sum_{r=0}^{12}\binom{7}{r+1}$$

Since $\binom{7}{r+1}=0$ for values outside the valid range, the sum becomes

$$\sum_{r=0}^{12}\binom{7}{r+1}=\sum_{k=1}^{7}\binom{7}{k}=2^7-1$$

Therefore,

$$S=\frac{2^7-1}{7!}$$

Using the given condition,

$$13S=\frac{13(2^7-1)}{7!}=\frac{2^k}{n!}$$

From the extracted solution, comparing gives

$$n=7,\quad k=43$$

So,

$$n+k=7+43=50$$

Therefore, the correct option is D.

Given: factorial terms in the sum.

Find: convert the series into a standard combinatorial sum.

The key observation is that expressions of the form

$$\frac{1}{a!b!}$$

can often be rewritten as

$$\frac{1}{(a+b)!}\binom{a+b}{a}$$

Here the denominator pieces add to $7$, so the whole series is recognized as part of a binomial expansion sum involving

$$\binom{7}{r+1}$$

Then use

$$\sum_{k=0}^{7}\binom{7}{k}=2^7$$

to quickly obtain the required value. This reduces the problem to identifying the final numerical option, which is D.