Let S = 1/(2·5!) + 1/(3!·23!) + 1/(5!·21!) + … up to 13 terms. If 13S = 2^k / n!, k ∈ ℕ, then n + k is equal to

Let S = 1/(1!·25!) + 1/(3!·23!) + 1/(5!·21!) + … up to 13 terms. If 13S = 2^k / n!, k ∈ ℕ, then n + k is equal to

Q. Let $$ S=\frac{1}{1!\cdot25!}+\frac{1}{3!\cdot23!}+\frac{1}{5!\cdot21!}+\cdots $$ up to 13 terms. If $$ 13S=\frac{2^k}{n!},\; k\in\mathbb{N}, $$ then $n+k$ is equal to

(A) 50

(B) 52

(D) 51

Correct Answer: 49

Explanation

The general term of the given series can be written as

$$ \frac{1}{(2r-1)!\,(27-2r)!}, \quad r=1,2,3,\ldots,13 $$

Using the identity

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

each term becomes

$$ \frac{1}{26!}\binom{26}{2r-1}. $$

Hence,

$$ S=\frac{1}{26!}\sum_{r=1}^{13}\binom{26}{2r-1}. $$

The sum of binomial coefficients with odd indices is

$$ \sum \binom{26}{\text{odd}}=2^{25}. $$

Therefore,

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

Now,

$$ 13S=\frac{13\cdot2^{25}}{26!}=\frac{2^{26}}{25!}. $$

Comparing with $\dfrac{2^k}{n!}$, we get

$$ k=26,\quad n=25. $$

Hence,

$$ n+k=25+26=\boxed{49}. $$

Explanation

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