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Show that: $$\begin{align*} &\sum_{k=0}^{n-1}\frac{\Bigl( (2n+1)(2n-1)(2n-3)(2n-5)\cdots (2n-2k+1)\Bigr)\sin x\;\cos^{2n-2k-1}x}{(2n+1)\Bigl( 2n(2n-2)(2n-4)\cdots(2n-2k)\Bigr)}\\ &\qquad= \sum_{k=1}^n\frac{\Bigl( (2n+1)(2n-1)(2n-3)(2n-5)\cdots(2k+1)\Bigr)\sin x\;\cos^{2k-1}x}{(2n+1)\Bigl(2n(2n-2)(2n-4)\cdots 2k\Bigr)}. \end{align*}$$

Thank you very much!

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Where did you get stuck? Is this homework? –  Inquest Mar 4 '12 at 8:42
Typos aside, the answer is: $k\to n-k$. –  Did Mar 4 '12 at 8:42
This is not homework! –  ᴊ ᴀ s ᴏ ɴ Mar 4 '12 at 8:52
@jason I see. I think that your question would have been better received if you had said that in the first place—you're not looking for a proof that this works, but simply an explanation of what is being done here. –  Tanner Swett Mar 4 '12 at 16:07
@DidierPiau, you could turn that comment into a fully fledged answer so that it can get accepted and this question put to rest :) –  Mariano Suárez-Alvarez Mar 5 '12 at 3:22

1 Answer 1

up vote 2 down vote accepted

The answer is: $k\to n-k$.

Edit Consider a sum of $n\geqslant1$ terms $S=\sum\limits_{k=0}^{n-1}x_{k}$. Then $S=\sum\limits_{i=1}^{n}x_{s(i)}$ for every bijection $s:\{1,2,\ldots,n\}\to\{0,2,\ldots,n-1\}$ because each term $x_k$ enters $S$ exactly once in both formulas. In particular, $S=\sum\limits_{i=1}^{n}x_{n-i}$.

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Why can we do like this? –  ᴊ ᴀ s ᴏ ɴ Mar 5 '12 at 9:58

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