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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)

$$\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $$

Community
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Quixotic
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  • You are probably missing terms in the left hand side... – Mariano Suárez-Álvarez Oct 25 '10 at 19:34
  • 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!) – Mariano Suárez-Álvarez Oct 25 '10 at 19:35
  • @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 ?! – Quixotic Oct 25 '10 at 19:38
  • @Debanjan: When I commented, you didn't have the "+..." and I checked it by having mathematica evaluate the expression for n=11. – Isaac Oct 25 '10 at 19:40
  • @ 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 :) – Quixotic Oct 25 '10 at 19:42
  • 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. – Isaac Oct 25 '10 at 19:46
  • 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. – Mariano Suárez-Álvarez Oct 25 '10 at 19:46
  • 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 :-) – Quixotic Oct 25 '10 at 19:51
  • 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. – Mariano Suárez-Álvarez Oct 25 '10 at 19:53
  • @Byron: No, $2^{n-1}$ is right. – Hans Lundmark Oct 25 '10 at 20:14
  • @Hans: Thanks, my fingers are faster than my brain.... –  Oct 25 '10 at 23:31
  • @Debanjan: are you preparing for IIT-JEE! –  Oct 26 '10 at 09:07

4 Answers4

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HINT:

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

  • +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 $ – Quixotic Oct 26 '10 at 02:57
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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.

Hans Lundmark
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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.

Yuval Filmus
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Seems easy is $n$ is odd. Multiplying both sides by $n!$ we have

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

leonbloy
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