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.
 A: In your sum, the term with $a=0$ is $0$. Take out the common factor $2^{n}$. And let $n=10$, for concreteness. So we want to find the sum 
$$S=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots +\frac{9}{2^{10}},$$
using a "general" method. Multiply $S$ by $2$. We get 
$$2S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots +\frac{9}{2^{9}}.$$
Write the expression for $S$ under the expression for $2S$, but pushed forward by one, so that the denominators line up nicely. Then the numerators are offset by $1$. Subtract. We get
$$2S-S=S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots +\frac{1}{2^9}-\frac{9}{2^{10}}.$$
Note that everything but the last term is a finite geometric series. I imagine you know how to find a closed form expression for that. If you don't, a simpler version of the same trick will do it. 
Now that you know what's going on, repeat for general $n$, and simplify to taste.  
Remark: $1.$ There was no need to get rid of the $2^n$. In fact things  look much nicer if you don't: No denominators! I am in retrospect very unhappy about getting rid of it. Repeat my argument using $T=1\cdot 2^n+2\cdot 2^{n-1}+3\cdot 2^{n-2}+\cdots +(n-1)\cdot 2^0$. It will quite a bit prettier.   
$2.$ Note that this is a distant relative of the so-called Gauss trick for finding the sum $1+2+3+\cdots +100$. The same idea works for a sum of shape $\sum_1^{n-1} kx^{k-1}$, if $x\ne 1$. You can adapt the idea to deal with $\sum_1^{n-1}k^2 x^k$, and other related sums. 
A: 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.
A: Proof it by induction, for $n=1$ it is true, than assume it is true for $n$ and show that this implies it is true for $n+1$
\begin{align*}
n+1+ \sum_{a=0}^n a \cdot 2^{n-a} &=n +1 +2 \cdot  \left(\sum_{a=0}^{n-1} a \cdot 2^{n-a-1}\right)+ n\\
&= 2 (2^n-1)+1\\
&=2^{n+1} -1
\end{align*}
A: Hint: Try writing the powers of two in binary, then sum up the terms starting with the least significant binary part and take a look at the carry-overs. Try some specific, small $n$ first.
