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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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I presume you aren't interested in the integral at all. – Aryabhata Feb 26 '12 at 18:06
No, I'm not interested in the integral at all! – Samuel Reid Feb 26 '12 at 18:12
up vote 2 down vote accepted

How about

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

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I hope I understood the question. – Aryabhata Feb 26 '12 at 18:07
Ahh! $n+k+1$, that's the term I was looking for in the exponent. Care to share how you came up with that? Or did it just seem obvious to you? – Samuel Reid Feb 26 '12 at 18:13
@Sam: The sign was dependent on both $n$ and $k$, in fact on the parity of $n+k$. If you increase $n$, it flips, if you increase $k$, it flips. So $n+k$ comes in. The +1 was to get the right sign. – Aryabhata Feb 26 '12 at 18:15
Excellent, that's exactly what I was looking for. :) – Samuel Reid Feb 26 '12 at 18:17
Each $I_n$ is an alternating sum of only $n-1$ terms. – Did Feb 26 '12 at 23:14

$$ \color{green}{I_n=\sum\limits_{i=2}^{n} (-1)^{n-i}\cdot i\cdot\left(\frac{\pi}{2} \right)^{i-1}} $$

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You should have edited the above answer. – vonbrand Sep 21 '15 at 18:59
@vonbrand What? – Did Sep 21 '15 at 20:45

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