Question
$$\frac{1}{{1!\, \cdot \left( {n - 1} \right)!}} + \frac{1}{{3!\, \cdot \left( {n - 3} \right)!}} + \frac{1}{{5!\, \cdot \left( {n - 5} \right)!}} + .....$$ is equal to
A.
$$\frac{{{2^{n - 1}}}}{{n!}}$$ for even values of $$n$$ only
B.
$$\frac{{{2^{n - 1}} + 1}}{{n!}} - 1$$ for odd values of $$n$$ only
C.
$$\frac{{{2^{n - 1}}}}{{n!}}$$ for all $$n \in N$$
D.
None of these
Answer :
$$\frac{{{2^{n - 1}}}}{{n!}}$$ for all $$n \in N$$
Solution :
Expression $$ = \frac{1}{{n!}}\left\{ {^n{C_1} + {\,^n}{C_3} + {\,^n}{C_5} + .....} \right\} = \frac{1}{{n!}} \cdot {2^{n - 1}}\,{\text{for all }}n \in N.$$