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I would like to know an idea on how to show this:

$\prod_{k=2}^{2n+1}$ $(1-{1\over k^2})$ = ${n+1\over 2n+1}$ $\forall$ $n\ge$ $2$.

I already checked for $2$ and tried it by induction but I didn't succeed.

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What have you tried so far? It is proper math.stackexchange etiquette to show that you have made an attempt to solve the problem. Also, please refrain in the future from adding irrelevant tags - this is not at all a calculus nor an abtract algebra problem. – Gyu Eun Lee Jan 16 '13 at 22:30
I try it for n+1 but I can´t write a lot in MathJax Sorry .I just want to know if there is another way to prove it. – Gmath Jan 16 '13 at 22:35
This is what's called a telescoping product. You can do this directly, without induction. Compare subsequent terms, carefully. – AndrewG Jan 16 '13 at 22:47

Two hints to start you off:

1) You do not need induction. The statement can be proved purely by algebraic manipulation of the given formula.

2) Write $$1 - \frac{1}{k^2} = \frac{k^2-1}{k^2}.$$ This can be put into the product easily. Think about factoring the terms here, and factorials in the product. Many terms will cancel.

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Telescopy is induction. sums and products are defined inductively so induction is inevitable. – Pedro Tamaroff Jan 16 '13 at 22:56
I guess if you want to be technical about it, then sure. I'm saying that there is no need to do a full proof by induction (i.e. test base cases, assume an induction hypothesis, etc.) here. – Gyu Eun Lee Jan 16 '13 at 22:59
Yes I agree is a semi induction but not fully thanks a lot to the both of you !!! – Gmath Jan 17 '13 at 3:43

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