Question

$\frac{1}{1!\cdot (n-1)!}+\frac{1}{3!\cdot (n-3)!}+\frac{1}{5!\cdot (n-5)!}+...+$is equal to

• A. $\frac{2^{n-1}}{n!}$ for even values of $n$ only
• B. $\frac{2^{n-1}+1}{n!}$ for odd values of $n$ only
• C. $\frac{2^{n-1}}{n!}$ for all $nϵN$
• D. none of these

Answer 1

The correct option is C $\frac{2^{n-1}}{n!}$ for all $nϵN$

For $n=1$ we get 1.
$=\frac{2^{1-1}}{1!}$
For $n=3$ we get
$\frac{1}{2!}+\frac{1}{3!}$$=\frac{4}{3!}$$=\frac{2^{3-1}}{3!}$
For $n=5$ we get
$\frac{1}{4!}+\frac{1}{3!2!}+\frac{1}{5!}$$=\frac{5+1}{5!}+\frac{1}{3!2!}$
$=\frac{6}{5!}+\frac{2\times 5}{5!}$
$=\frac{16}{5!}$
$=\frac{2^{5-1}}{5!}$
Hence for $n=2k+1$ where K belongs to whole numbers, we get
$\frac{2^{2k}}{(2k+1)!}$$=\frac{2^{n-1}}{n!}$