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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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    You are probably missing terms in the left hand side...2010-10-25
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    In any case, exactly the same ideas used to solve your last few questions apply here. Why don't you instead explain to us what you tried to do and we can help you with it? (Otherwise, you recent stream of problems feels way too close to our doing your list of problems on identities with binomial coefficients!)2010-10-25
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    @Isaac♦ :I haven't checked it with 11 but Would you please tell me how did you computed the value for n = 11 so fast ? ;) I guess you are using some tools ?!2010-10-25
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    @Debanjan: When I commented, you didn't have the "+..." and I checked it by having mathematica evaluate the expression for n=11.2010-10-25
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    @ Mariano Suárez-Alvarez : No,I choose to disagree these are not what you think and I know many people who would rather cram them up instead of trying to proof them some other way around except the trivial induction.Actually according to exam needs it will be good with just induction (since I am only supposed to click the correct option ) but I noticed that I am learning more while trying to solve them the other way :)2010-10-25
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    My instinct here is that doing something with a common denominator on the left side would be helpful, but I haven't actually tried it.2010-10-25
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    Trust me: you would learn **much** more by dpoing this problem yourself, using induction, or whatever other method you can come up. Almost for each answer of your last question, you can come up with an solution using the same idea for this one.2010-10-25
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    I don't really know why do you think that I haven't tried myself before typing the entire thing here ?! But I would never know about those tricks that Isaac and svenkatr pointed out in the earlier threads,so I guess I am learning more by discussing here :-)2010-10-25
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    As I said, most of the *same* tricks that solved http://math.stackexchange.com/questions/7757/how-to-prove-this-binomial-identity-sum-r0n-r-n-choose-r-n2n-1 apply verbatim here. If you learned about them there, you could apply them here.2010-10-25
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    @Byron: No, $2^{n-1}$ is right.2010-10-25
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    @Hans: Thanks, my fingers are faster than my brain....2010-10-25
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    @Debanjan: are you preparing for IIT-JEE!2010-10-26

4 Answers 4

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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

$$ {n \choose 1} + {n \choose 3} + ... {n \choose n} = \frac{1}{2} \left[ {n \choose 0} + {n \choose 1} + ... {n \choose n} \right] = 2^{n-1}$$