Question

Find the value of $\frac{1}{(n-1)!}+\frac{1}{(n-3)!3!}+\frac{1}{(n-5)!5!}+...$

Answer 1

$\because 1!=1$

$\therefore$ The given series can be written as

$\frac{1}{(n-1)!1!}+\frac{1}{(n-3)!3!}+\frac{1}{(n-5)!5!}+........+n$

$\because$ sum of values of each terms in fraction are equal

i.e., $(n-1)+1=(n-3)+3=(n-5)+5=........$

From (1)

$\frac{1}{n!}[\frac{n!}{(n-1)!1!}+\frac{n!}{(n-3)!3!}+\frac{n!}{(n-5)!5!}+........]$

$=\frac{1}{n!}\left({}^n{C}_1{+}^n{C}_3{+}^n{C}_5+......\right)=\frac{2^{n-1}}{n!}$