Closed form of this arithmetic series? I am computing the partial sums $s_{nn}$ of a series given by 
$a_{ij} = {1 \over 2}^{j-i}$ if $j>i$, $a_{ij} = -1$ if $j=i$ and $a_{ij} = 0$ if $j < i$
The $s_{nn}$ is the partial sum $\sum_{i,j =1}^n a_{ij}$. I computed that $$s_{nn}=\sum_{k=1}^{n-1} (n-k) {1 \over 2^k}$$
and now I am trying to find a closed formula for $s_{nn}$ but I'm stuck. Please can someone help me how to do this? 

Assuming your expression $s_{nn} = \sum_{k=1}^{n-1} (n-k) {1 \over 2^k}$ is correct, I work from there.
$$s_{nn} = \sum_{k=1}^{n-1} (n-k) {1 \over 2^k} = n \sum_{k=1}^{n-1}{1 \over 2^k} - \sum_{k=1}^{n-1}{k \over 2^k}$$
As the first term is clearly a geometric series, I focus on the latter term.
$$ \sum_{k=1}^{n-1}{k \over 2^k} = \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + \cdot \cdot \cdot + \frac{n-1}{2^{n-1}} $$
$$ \sum_{k=1}^{n-1}{k \over 2^k} = (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdot \cdot \cdot + \frac{1}{2^{n-1}}) + (\frac{1}{4} + \frac{1}{8} + \cdot \cdot \cdot + \frac{1}{2^{n-1}}) + (\frac{1}{8} + \cdot \cdot \cdot + \frac{1}{2^{n-1}}) + \cdot \cdot \cdot + \frac{1}{2^{n-1}} $$
Which is a sum of geometric series. Plugging in the closed form expression for each individual series will result in another geometric series, finally providing the partial sum.
