Which is the "fastest" way to compute $\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $? 
I am looking for the "fastest" paper-pencil approach to compute $$\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} $$

This is a quantitative aptitude problem and the correct/required answer is $3.75$
In addition, I am also interested to know how to derive a closed form for an arbitrary $n$ using mathematica I got $$\sum \limits_{i=1}^{n} \frac{10i-5}{2^{i+2}} =  \frac{5 \times  \left(3 \times 2^n-2 n-3\right)}{2^{n+2}}$$
Thanks,

First note that $\displaystyle\frac{10i-5}{2^{i+2}} = \frac{5(2i-1)}{2^{i+2}}$
Secondly, using the sum of a geometric series you can show that $2^0 + 2^1 + 2^2 + ... + 2^k = 2^{k+1}-1$.
$\displaystyle \begin{align*} \text{So } \sum \limits_{i=1}^{n} \frac{10i-5}{2^{i+2}} 

&= 5(\frac{2\times1-1}{2^{1+2}}+\frac{2\times2-1}{2^{2+2}}+\frac{2\times3-1}{2^{3+2}}+...+\frac{2\times n-1}{2^{n+2}})\\

&= 5(\frac{2^{n-1}(2\times1-1)+2^{n-2}(2\times2-1)+2^{n-3}(2\times3-1)+...+2^{0}(2\times n-1)}{2^{n+2}}) \\

&= 5(\frac{2(2^{n-1}\times1+2^{n-2}\times2+2^{n-3}\times3+...+2^{0}\times n)-(2^{n-1}+2^{n-2}+2^{n-3}+...+1)}{2^{n+2}})\\

&= 5(\frac{2(2^{n-1}\times1+2^{n-2}\times2+2^{n-3}\times3+...+2^{0}\times n)-(2^n-1)}{2^{n+2}})\\

&= 5(\frac{2((2^{n-1}+2^{n-2}+2^{n-3}+...+1)+(2^{n-2}+2^{n-3}+2^{n-4}+...+1)+...+(2+1)+(1))-(2^n-1)}{2^{n+2}})\\

&= 5(\frac{2((2^{n}-1)+(2^{n-1}-1)+(2^{n-2}-1)+...+(2^{2}-1)+(2-1))-(2^n-1)}{2^{n+2}})\\

&= 5(\frac{2((2^{n}+2^{n-1}+2^{n-2}+...+2) + (-1)\times n)-(2^n-1)}{2^{n+2}})\\

&= 5(\frac{2(2\times (2^n - 1))- n)-(2^n-1)}{2^{n+2}})\\

&= 5(\frac{4 \times 2^n - 4 - 2n -2^n + 1}{2^{n+2}})\\

&= 5(\frac{3 \times 2^n - 2n - 3}{2^{n+2}})\text{ which is the closed form you got from Mathematica}\end{align*}$
