Binomial Theorem Part 3

Concept:The given series can be expressed as a sum of binomial coefficients using the factorial form of $\binom{n}{r}$.Explanation:Multiply the sum by $\frac{26!}{26!}$ to rewrite each term as a binomial coefficient.$S = \frac{1}{26!} \left[ \frac{26!}{1!25!} + \frac{26!}{3!23!} + \frac{26!}{5!21!} + \dots \right]$.This becomes $S = \frac{1}{26!} \left[ \binom{26}{1} + \binom{26}{3} + \binom{26}{5} + \dots \right]$ with 13 terms.The subscripts follow an arithmetic progression: $1, 3, 5, \dots, 25$ (since $1 + (13-1)\cdot2 = 25$).The sum of odd-indexed binomial coefficients is $2^{n-1}$ for $n>0$. Here $n=26$, so the sum is $2^{25}$.Thus $S = \frac{2^{25}}{26!}$.Given $13S = \frac{2^k}{n!}$, substitute $S$: $13 \cdot \frac{2^{25}}{26!} = \frac{2^k}{n!}$.Simplify $26! = 26 \cdot 25!$, so $\frac{13 \cdot 2^{25}}{26 \cdot 25!} = \frac{2^{24}}{25!} = \frac{2^k}{n!}$.Hence $n = 25$ and $k = 24$.Answer:$n + k = 25 + 24 = 49$.Thus the correct option is C. 49.