Let S=11!,25!+13!,23!+15!,21!+ up to 13 terms.

Question

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

Accepted Answer

Multiply numerator and denominator of each term by $26!$:

$$S=\frac{1}{26!}\left[\binom{26}{1}+\binom{26}{3}+\binom{26}{5}+\cdots\text{(13 terms)}\right]$$

The lower indices $1,3,5,\dots$ form an A.P. with common difference 2; the 13th term is $1+(13-1)\times2=25$. So

$$S=\frac{1}{26!}\left[\binom{26}{1}+\binom{26}{3}+\cdots+\binom{26}{25}\right]=\frac{1}{26!}\cdot 2^{25}$$

(using $\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\cdots=2^{n-1}$).

So $13S = \dfrac{13\cdot 2^{25}}{26!}=\dfrac{13\cdot2^{25}}{26\cdot25!}=\dfrac{2^{24}}{25!}$.

Comparing with $\dfrac{2^k}{n!}$: $n=25,\ k=24$, so $n+k=49$.

Solutions · 1