Question

Evaluate the series \[ \frac{1}{25!}+\frac{1}{3!\,23!}+\frac{1}{5!\,21!}+\cdots \text{ up to 13 terms.} \]

• A. $\dfrac{2^{26}}{26!}$
• B. $\dfrac{2^{25}}{26!}$
• C. $\dfrac{2^{26}}{25!}$
• D. $\dfrac{2^{25}}{25!}$

Hint:

Whenever factorial products appear, try converting them into binomial coefficients to simplify series efficiently.

Answer 1

The Correct Option is B

Step 1: Write the general term of the series.
The given series can be written as: \[ \sum_{k=0}^{12} \frac{1}{(2k+1)!\,(25-2k)!} \] Step 2: Use the binomial coefficient identity.
Recall: \[ \binom{26}{r}=\frac{26!}{r!(26-r)!} \] Hence, \[ \frac{1}{(2k+1)!(25-2k)!} =\frac{26}{26!}\binom{26}{2k+1} \] Step 3: Rewrite the series.
\[ \sum_{k=0}^{12} \frac{1}{(2k+1)!(25-2k)!} =\frac{26}{26!}\sum_{k=0}^{12}\binom{26}{2k+1} \] Step 4: Use the binomial sum identity.
The sum of odd binomial coefficients is: \[ \sum_{\text{odd }r}\binom{26}{r}=2^{25} \] Step 5: Final calculation.
\[ \text{Sum}=\frac{26}{26!}\cdot 2^{25} =\frac{2^{25}}{26!} \]