Finding a simple expression for this series expansion without a piecewise definition 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?

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