An approximate answer
We'll split the sum you're looking for up into
$ 10 \sum_{i=1}^{10} {i \over 2^{i+2}} - 5 \sum_{i=1}^{10} {1 \over 2^{i+2}}. $
Call this $10S_1 - 5S_2$.
We can write $ S_1 = \sum_{i=1}^{10} {i \over 2^{i+2}} = {1 \over 4} \sum_{i=1}^{10} {i \over 2^i} $ and the result $\sum_{i=1}^\infty i/2^i = 2$ is well-known; thus $S_1 \approx 1/2$.
Similarly, $S_2 \approx \sum_{i=1}^\infty 1/2^{i+2} = 1/4$ by the usual sum of a geometric series.
So your sum is approximately $15/4$. In fact the infinite sum
$ \sum_{i=1}^\infty {10i-5 \over 2^{i+2}} $
is exactly 15/4. We've left off some small positive terms so your sum is a bit less than $15/4$.
An exact form for the sum
As for getting an exact form for the sum: call it $f(n)$. Then we have
$ f(n) = {15 \over 4} - \left( 10 \sum_{i=n+1}^\infty {i \over 2^{i+2}} - 5 \sum_{i=n+1}^\infty {1 \over 2^{i+2}} \right). $
Write this as $f(n) = 15/4 - g(n) + h(n)$.
$h(n)$ is easy -- it's the sum of a geometric series with its first term $5/2^{n+3}$ and ratio $1/2$, so $h(n) = 5/2^{n+2}$.
$g(n)$ is a bit harder. So that we don't have so many constants floating around, consider
$ G(n) = \sum_{i={n+1}}^\infty {i \over 2^i} $
and you can see $g(n) = (5/2) G(n)$. Now you can write
$ G(n) = {(n+1) \over 2^{n+1}} + {{n+2} \over 2^{n+2}} + {{n+3} \over 2^{n+3}} + \cdots $
and this is just
$ G(n) = \left( {n \over 2^{n+1}} + {n \over 2^{n+2}} + {n \over 2^{n+3}} + \cdots \right) + {1 \over 2^n} \left( {1 \over 2^1} + {2 \over 2^2} + {3 \over 2^3} + \cdots \right). $
The first sum is a geometric series, summing to $n/2^n$; the second sum is $2$. Thus $G(n) = (n+1)/2^n$. Therefore you get
$ f(n) = {15 \over 4} - {5 \over 2} {(n+2) \over 2^n} + {5 \over 2^{n+2}}. $
A sum that everybody should know, but lots of people don't
Finally, I used the result $\sum_{i=1}^\infty i/2^i$ twice here, both in getting the approximation and in getting the exact form. How can we prove that? One way is to write $ {1 \over 2} + {2 \over 4} + {3 \over 8} + {4 \over 16} + \cdots $ as a sum of one $1/2$, two $1/4$s, three $1/8$s, and so on; then regroup those terms as $ \left( {1 \over 2} + {1 \over 4} + {1 \over 8} + \cdots \right) + \left( {1 \over 4} + {1 \over 8} + {1 \over 16} + \cdots \right) + \left( {1 \over 8} + {1 \over 16} + {1 \over 32} + \cdots \right) + \cdots $ Each pair of parentheses contains a geometric series; summing those gives $1 + 1/2 + 1/4 + 1/8 + \cdots$, another geometric series, which has sum $2$.
Alternatively, note that $ {1 \over 1-z} = 1 + z + z^2 + z^3 + \cdots $ and differentiating both sides gives $ {1 \over (1-z)^2} = 1 + 2z + 3z^2 + \cdots $. Multiply both sides by $z$ to get $ {z \over (1-z)^2} = z + 2z^2 + 3z^3 + \cdots $ and plug in $z = 1/2$ to get $ {1/2 \over (1-1/2)^2} = {1 \over 2} + {2 \over 2^2} + {3 \over 2^3} + \cdots. $