Showing $\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $ I am stuck while proving this identity I verified it using induction but like the other two (1) (2), I am seeking a more of a general way (algebraic will be much appreciated)
$$\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!}  $$

Seems easy is $n$ is odd. Multiplying both sides by $n!$ we have
$$ {n \choose 1} + {n \choose 3} + ...  {n \choose n} = \frac{1}{2} \left[  {n \choose 0} + {n \choose 1} + ... {n \choose n}  \right]  = 2^{n-1}$$
