Summation Simplification I am attempting to solve a problem that I posed myself, but I can't figure out how to simplify the solution from the "messy" state in which it currently exists. My mathematical background does not yet contain calculus or related studies, so it is quite likely there is a simple formula or theorem that can be applied that I do not know. I know the solution (because of Wolfram Alpha), but I am interested in determining how to solve it, and similar problems as they present themselves, by myself. The expression is:
$n+\displaystyle\sum\limits_{a=0}^{n-1}a\cdot2^{n-a-1}$
This is equal to $2^n-1$, according to Wolfram Alpha. How do I arrive at that solution? I greatly appreciate any help you can provide me on this problem. Thank you for your time and for any assistance you can give.

You can solve it by splitting the sum and applying the formula for the geometric series
$$\sum_{i=0}^n q^i = \frac{q^{n+1}-1}{q-1}$$
with $q=2$ two times on the result:
\begin{align}
\sum_{a=1}^{n-1} a 2^{n-a-1} &= 1\cdot 2^{n-2} + 2\cdot 2^{n-3} + \cdots 
+ (n-1) 2^0 \\
&=(2^{n-2}  + \cdots + 2^0) 
+ ( 2^{n-3} + \cdots + 2^0)
+ (2^{n-4}  \cdots + 2^0) + \cdots + (2^0)\\
&=(2^{n-1} -1) + (2^{n-2} -1) + \cdots + (2^1 - 1)\\
&=(2^{n-1} + 2^{n-2} + \cdots + 2^1) - (n-1)\\
&=(2^{n} - 2) - (n-1)\\
&=2^{n} - 1 - n\\
\end{align}
Now add $n$ to both sides.
