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I am doing some practice Calculus questions and I ran into the following problem which ended up having a reduction formula with a neat expansion that I was wondering how to express in terms of a series. Here it is: consider $ I_{n} = \int_{0}^{\pi /2} x^n \sin(x) dx $ I obtained the reduction formula $ I_{n} = n\left(\frac{\pi}{2}\right)^{n-1} - n I_{n-1}. $ I started incorrectly computing up to $I_{6}$ with the reduction formula $ I_{n} = n\left(\frac{\pi}{2}\right)^{n-1} - I_{n-1} $ by accident which ended up having a way more interesting pattern than the correct reduction formula. So, after computing $I_{0} = 1$, the incorrect reduction expansion was, $ I_{1} = 0 \\ I_{2} = \pi \\ I_{3} = \frac{3\pi^2}{2^2} - \pi \\ I_{4} = \frac{4\pi^3}{2^3} - \frac{3\pi^2}{2^2} + \pi \\ I_{5} = \frac{5\pi^4}{2^4} - \frac{4\pi^3}{2^3} + \frac{3\pi^2}{2^2} - \pi \\ I_{6} = \frac{6\pi^5}{2^5} - \frac{5\pi^4}{2^4} + \frac{4\pi^3}{2^3} - \frac{3\pi^2}{2^2} + \pi \\ $ Note that $\pi = \frac{2\pi}{2^1}$, of course, which stays in the spirit of the pattern. How could I give a general expression for this series without defining a piecewise function for the odd and even cases? I was thinking of having a term in the summand with $(-1)^{2i+1}$ or $(-1)^{2i}$ depending on it was a term with an even or odd power for $n$, but that led to a piecewise defined function. I think that it will look something like the following, where $f(x)$ is some function that handles which term gets a negative or positive sign depending on whether $n$ is an even or odd power in that term: $\sum\limits_{i=1}^{n} n \left(\frac{\pi}{2} \right)^{n-1} f(x)$

Any ideas on how to come up with a general expression for this series?

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    No, I'm not interested in the integral at all!2012-02-26

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How about

$\sum_{k=1}^{n} k \left(\frac{\pi}{2}\right)^{k-1} \cdot (-1)^{n+k+1}$

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    @DidierPiau: I agree :-)2012-02-27
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$ \color{green}{I_n=\sum\limits_{i=2}^{n} (-1)^{n-i}\cdot i\cdot\left(\frac{\pi}{2} \right)^{i-1}} $

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    @vonbrand What?2015-09-21