Let S=(1)/(25!)+(1)/(3!23!)+(1)/(5!21!)+ up to 13 terms.

Question

Let $S=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots$ up to 13 terms. If $13S=\frac{{2}^{k}}{n!},k\in N$, then $n+k$ is equal to

Accepted Answer

$S=\frac{1}{1!25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots 13$

multiply in numerator and denominator by 26 !

$\begin{gathered} S=\frac{1}{26!}[\frac{26!}{1!25!}+\frac{26!}{3!23!}+\frac{26!}{5!2!}+\ldots 13] \ S=\frac{1}{26!}[{}^{26}{C}_{1}+{}^{26}{C}_{3}+{}^{26}{C}_{5}+\ldots 13],{{}^{n}{C}_{r}=\frac{n!}{(n-r)!r!}} \end{gathered}$

Lets find 13th term lower subscript are in A.P with, common difference $=2$

$\begin{gathered} {T}_{13}=1+(13-1)2=25 \ S=\frac{1}{26!}[{}^{26}{C}_{1}+{}^{26}{C}_{3}+{}^{26}{C}_{5}+\ldots {}^{26}{C}_{25}] \ S=\frac{1}{26!}[{2}^{26-1}],{{}^{n}{C}_{1}+{}^{n}{C}_{3}+{}^{n}{C}_{5}\cdots ={2}^{n-1}} \end{gathered}$

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

It is given that $13S=\frac{{2}^{K}}{n!}$

substitute value of $S$

$\begin{gathered} 13	imes \frac{{2}^{25}}{26!}=\frac{{2}^{K}}{n!} \ \frac{13	imes {2}^{25}}{26	imes 25!}=\frac{{2}^{K}}{n!} \ \frac{{2}^{24}}{25!}=\frac{{2}^{K}}{n!} \end{gathered}$

so $n=25,K=24$

$n+K=25+24=49$

Solutions · 1