The problem asks us to find the value of $n+k$ given a series $S$ and an equation relating $S$ to powers and factorials.
Series Sum Calculation
The given series is $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ and it has 13 terms.
Let's analyze the terms:
- Term 1: $\frac{1}{25!}$
- Term 2: $\frac{1}{3!23!}$ (Sum of factorials = $3+23=26$)
- Term 3: $\frac{1}{5!21!}$ (Sum of factorials = $5+21=26$)
The structure suggests a connection to binomial coefficients, $\binom{n}{r} = \frac{n!}{r!(n-r)!}$. The sum of the numbers in the factorials for the second and third terms is 26. The series likely involves terms of the form $\frac{1}{p!q!}$ where $p+q=26$ and $p$ takes odd values.
The odd numbers from 1 to 25 are $1, 3, 5, \dots, 25$. The count is $\frac{25-1}{2} + 1 = 13$. This matches the number of terms given.
Assuming the series follows this pattern, the first term $\frac{1}{25!}$ is likely intended to be $\frac{1}{1!25!}$. The series becomes:
$S = \frac{1}{1!25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots + \frac{1}{25!1!}$This can be written as:
$S = \sum_{i=0}^{12} \frac{1}{(2i+1)!(26-(2i+1))!} = \sum_{i=0}^{12} \frac{1}{(2i+1)!(25-2i)!}$Applying Binomial Identity
To relate this to binomial coefficients, multiply $S$ by $26!$:
$26! S = \sum_{i=0}^{12} \frac{26!}{(2i+1)!(26-(2i+1))!}$ $26! S = \sum_{i=0}^{12} \binom{26}{2i+1}$This sum represents the sum of binomial coefficients with odd lower indices:
$26! S = \binom{26}{1} + \binom{26}{3} + \binom{26}{5} + \dots + \binom{26}{25}$Using the identity $\sum_{r \text{ odd}} \binom{n}{r} = 2^{n-1}$, with $n=26$:
$26! S = 2^{26-1} = 2^{25}$Therefore, the sum of the series is:
$S = \frac{2^{25}}{26!}$Finding n and k
We are given the equation $13S = \frac{2^k}{n!}$.
Substitute the value of $S$ we found:
$13 \times \frac{2^{25}}{26!} = \frac{2^k}{n!}$Rewrite $26!$ as $26 \times 25!$:
$\frac{13 \times 2^{25}}{26 \times 25!} = \frac{2^k}{n!}$Simplify the left side:
$\frac{13 \times 2^{25}}{2 \times 13 \times 25!} = \frac{2^k}{n!}$ $\frac{2^{24}}{25!} = \frac{2^k}{n!}$By comparing the terms, we can identify $n$ and $k$:
$n = 25$ $k = 24$Calculating n + k
The question asks for the value of $n+k$.
$n + k = 25 + 24 = 49$Note: The derived result is 49. This differs from the provided answer option A (52). The solution follows the standard mathematical interpretation of the series provided.
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