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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)!} $

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    @Debanjan: are you preparing for IIT-JEE!2010-10-26

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HINT:

$(1+x)^{n} - (1-x)^{n} = 2 \Biggl[ {n \choose 1} x + {n \choose 3} x^{3} + \cdots \Biggr]$

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    +1,Thanks,this is quite sufficient for me, also I do realized that the same results hold for $ \frac {1}{(n)!} + \frac {1}{2!(n-2)!} + \frac {1}{4!(n-4)!} +\frac {1}{6!(n-6)!} + \cdots $2010-10-26
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For even $n$, look at the coefficient of $x^n$ in the expansion of $\sinh^2 x = \frac12(\cosh 2x-1)$. For odd $n$, modify this idea suitably.

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Multiplying by $n!$, the left-hand side it the number of subsets of $[n]$ of odd size, and the right-hand side is $2^{n-1}$. This is because for each set $S \subseteq [n-1]$, exactly one of $S,S \cup {n}$ has odd size.

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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}$