A) \[\frac{{{2}^{n}}}{n!}\]; for all even values of n
B) \[\frac{{{2}^{n-1}}}{n!}\]; for all values of n i.e., all even odd values
C) 0
D) None of these
Correct Answer: B
Solution :
Multiplying each term by n! the question reduces to \[\frac{n!}{1!(n-1)!}+\frac{1}{3!}.\frac{n!}{(n-3)\,!}+\frac{1}{5!}.\frac{n!}{(n-5)!}+....\] \[={{\,}^{n}}{{C}_{1}}+{{\,}^{n}}{{C}_{3}}+{{\,}^{n}}{{C}_{5}}+....={{2}^{n-1}}\]. Thus\[\frac{1}{1!(n-1)!}+\frac{1}{3!(n-3)!}+\frac{1}{5!(n-5)!}+....\]\[=\frac{1}{n!}{{2}^{n-1}}\].
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