Prove for all whole numbers n, $(n+1)(n+2)...(2n-1)(2n)=2^n(1)(3)(5)...(2n-1)$.
I got upto $(2n)(2n-2)(2n-4)...=2^n$, after which I'm stuck.
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1How do you get that as $2^n$? It should be $2^n(n)(n-1)(n-2)\ldots (3)(2)(1)=2^nn!$ – Adam Hughes Sep 09 '14 at 02:04
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I find this easier to read as $\prod_{i=1}^n (n+i) = 2^n \prod_{i=1}^n (2i-1)$, just because this notation says more about how the factors are related. – NoName Sep 09 '14 at 02:05
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@ZachGershkoff not a sum. – Adam Hughes Sep 09 '14 at 02:08
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@AdamHughes : This is definitely a duplicate. – Patrick Da Silva Sep 09 '14 at 02:09
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1One must obviously proceed by induction. – Sep 09 '14 at 02:20
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2$$\begin{align}RHS&=2^n(1\cdot 3\cdot 5\cdots (2n-1))\&=\frac {2^n[1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdots (2n-1)(2n)]}{2\cdot 4\cdot 6\cdots (2n)}\&=\frac {2^n[1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdots (2n-1)(2n)]}{2^n[1\cdot 2\cdot 3\cdots (n)]}\&=\frac {2^n(2n)!}{2^n(n!)}\&=\frac {(2n)!}{n!}\&=(n+1)(n+2)(n+3)\cdots (2n-1)(2n)=LHS\qquad \blacksquare\end{align}$$ – Hypergeometricx Sep 09 '14 at 15:52