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,
 A: 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:
$$
\begin{eqnarray}
  \sum_{k=1}^n \frac{10 k -5}{2^{k+2}} &=& \frac{5}{4} \sum_{k=1}^n (2k-1)\left(\frac{1}{2}\right)^k  \\
   &=& \frac{5}{4} \frac{1}{2} \left( -(2^{1-n} - 2) +  (n-1) 2^{2-n}- n 2^{1-n} + 4  \right) \\
   &=& \frac{5}{4} \left(  3 + (n-3) 2^{-n}    \right) 
\end{eqnarray}
$$  
A: 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*}$
A: The sum $S= \sum \limits_{i=1}^n \frac{1}{2^i}=1-\frac{1}{2^n}$ is geometric, thus easy to calculate.
Here is a simple elementary way of calculating 
$$T=\sum_{i=1}^n \frac{i}{2^i} \,.$$:
$$T=\sum_{i=1}^n \frac{i}{2^i} =\frac{1}{2}+ \sum_{i=2}^n \frac{i}{2^i} =\frac{1}{2}-\frac{n+1}{2^{n+1}}+ \sum_{i=2}^{n+1} \frac{i}{2^i}
\,.$$
Changing the index in the last sum yields:
$$T= \frac{2^n-n-1}{2^{n+1}}+\sum_{i=1}^{n} \frac{i+1}{2^{i+1}}=\frac{2^n-n-1}{2^{n+1}}+\sum_{i=1}^{n} \frac{i}{2^{i+1}}+\sum_{i=1}^{n} \frac{1}{2^{i+1}}    \,.$$
Thus
$$T= \frac{2^n-n-1}{2^{n+1}}+\frac{1}{2}T+ \frac{1}{2}-\frac{1}{2^{n+1}}$$
Thus
$$\frac{1}{2}T=\frac{2^{n+1}-n-2}{2^{n+1}}\,.$$
Hence
$$\sum_{i=1}^n \frac{i}{2^i} =\frac{2^{n+1}-n-2}{2^{n}} \,.$$
A: $\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} =\frac{10}{4} \sum_{i=1}^{10} \frac{i}{2^i}-\frac{5}{4}\sum_{i=1}^{10} \frac{1}{2^i}$
Second $\sum$ is obviously $1-\frac{1}{1024}$.
First $\sum$:
$\frac{1}{2} +$
$\frac{1}{4} + \frac{1}{4} +$
$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} +$
$\vdots$
$\frac{1}{1024} + \frac{1}{1024} + \dots + \frac{1}{1024}$
Summing by columns, $(1-\frac{1}{1024})+(\frac{1}{2}-\frac{1}{1024})+\dots+(\frac{1}{512}-\frac{1}{1024})=(1+\frac{1}{2}+\dots+\frac{1}{512})-\frac{10}{1024}=\frac{1023}{512}-\frac{10}{1024}$
Rest is boring and easy.
