JEE Mathematics 2023 Question with Solution

Given:

$$\sum_{r=1}^{26} \frac{1}{(2r-1)!\,(51-(2r-1))!}$$

Find: The value of the given sum and hence the correct option.

Rewrite each term using binomial coefficients:

$$\frac{1}{(2r-1)!\,(51-(2r-1))!} = \binom{51}{2r-1}\frac{1}{51!}$$

Therefore,

$$\sum_{r=1}^{26} \frac{1}{(2r-1)!\,(51-(2r-1))!} = \sum_{r=1}^{26} \binom{51}{2r-1}\frac{1}{51!}$$ $$= \frac{1}{51!}\sum_{r=1}^{26} \binom{51}{2r-1}$$

Now the sum of all odd binomial coefficients for $51$ is

$$\binom{51}{1}+\binom{51}{3}+\cdots+\binom{51}{51}=2^{50}$$

Hence,

$$\frac{1}{51!}\sum_{r=1}^{26} \binom{51}{2r-1}=\frac{2^{50}}{51!}$$

Therefore, the correct value is $\frac{2^{50}}{51!}$, so the correct option is B. The solution labels it as option D, but the computed value matches option B in the given options.

Given: The terms have factorials whose sum in the denominator is always $51!$ after introducing binomial coefficients.

Find: The sum quickly.

Observe that

$$\frac{1}{(2r-1)!\,(51-(2r-1))!}=\frac{1}{51!}\binom{51}{2r-1}$$

So the whole series is

$$\frac{1}{51!}\left(\binom{51}{1}+\binom{51}{3}+\cdots+\binom{51}{51}\right)$$

Using the identity that the sum of odd coefficients in $(1+1)^{51}$ is $2^{50}$, we get

$$\frac{1}{51!}\cdot 2^{50}=\frac{2^{50}}{51!}$$

Therefore, the correct option is B.