Question

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

• A. $ \frac{{{2}^{n}}}{n!} $
• B. $ \frac{{{2}^{n-1}}}{n!} $
• C. 0
• D. none of these

Answer 1

Answer: (B)

$ \frac{1}{1!(n-1)!}+\frac{1}{3!(n-3)!}+\frac{1}{5!(n-5)!}+.... $
Multiplying numerator and denominator by $ n! $
$ =\frac{1}{n!}\left[ \frac{n!}{1!(n-1)!}+\frac{n!}{3!(n-3)!}+\frac{n!}{5!(n-5)!}+... \right] $
$ =\frac{1}{n!}\left[ {{\,}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{3}}{{+}^{n}}{{C}_{5}}+... \right] $
$ =\frac{1}{n!}{{2}^{n-1}} $