Question

Let \[ S=\frac{1}{2!5!}+\frac{1}{3!2!3!}+\frac{1}{5!2!1!}+\cdots \text{ up to 13 terms}. \] If $13S=\dfrac{2^k}{n!}$, $k\in\mathbb{N}$, then $n+k$ is equal to

• A. 52
• B. 51
• C. 49
• D. 50

Hint:

Whenever factorial sums appear, try rewriting them using binomial coefficients.

Answer 1

The Correct Option is D

Step 1: Observing the pattern.
Each term is of the form \[ \frac{1}{(r+1)!\,(6-r)!},\quad r=0,1,2,\ldots,12 \] Thus, \[ S=\sum_{r=0}^{12}\frac{1}{(r+1)!(6-r)!} \] Step 2: Writing in binomial coefficient form.
\[ \frac{1}{(r+1)!(6-r)!}=\frac{1}{7!}\binom{7}{r+1} \] Hence, \[ S=\frac{1}{7!}\sum_{r=0}^{12}\binom{7}{r+1} \] Step 3: Evaluating the sum.
\[ \sum_{r=0}^{12}\binom{7}{r+1}=\sum_{k=1}^{7}\binom{7}{k}=2^7-1 \] \[ S=\frac{2^7-1}{7!} \] Step 4: Using the given condition.
\[ 13S=\frac{13(2^7-1)}{7!}=\frac{2^k}{n!} \] Comparing factorial and power of 2, \[ n=7,\quad k=43 \] \[ n+k=7+43=50 \]