<s>[INST] 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,
 [/INST] I would do it like this. Using $x \frac{\mathrm{d}}{\mathrm{d} x}\left( x^k \right) = k x^k$, and $\sum_{k=1}^n x^k = x \frac{x^n-1}{x-1}$. Then
$$\begin{eqnarray}
   \sum_{k=1}^n \left( a k +b\right) x^k &=& \left( a x \frac{\mathrm{d}}{\mathrm{d} x} + b\right) \circ \sum_{k=1}^n x^k = \left( a x \frac{\mathrm{d}}{\mathrm{d} x} + b\right) \circ \left( x \frac{x^n-1}{x-1} \right) \\
    &=& x \left( a x \frac{\mathrm{d}}{\mathrm{d} x} + a + b\right) \circ \left(\frac{x^n-1}{x-1} \right) \\ 
    &=& x \left(  (a+b) \frac{x^n-1}{x-1} + a x \frac{ n x^{n-1}(x-1) - (x^n-1) }{(x-1)^2}  \right) \\ 
 &=& x \left(  (a+b) \frac{x^n-1}{x-1} + a x \frac{ (n-1) x^{n} - n x^{n-1} + 1 }{(x-1)^2}  \right)
\end{eqnarray}
$$
Now applying this:
$$
\begin{eqnarray}
  \sum_{k=1}^n \frac{10 k -5}{2^{k+2}} &=& \frac{5}{4} \sum_{k=1}^n (2k-1)\left(\frac{1}{2}\right)^k  \\
   &=& \frac{5}{4} \frac{1}{2} \left( -(2^{1-n} - 2) +  (n-1) 2^{2-n}- n 2^{1-n} + 4  \right) \\
   &=& \frac{5}{4} \left(  3 + (n-3) 2^{-n}    \right) 
\end{eqnarray}
$$  
</s>