<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] 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
$$\sum_{i=1}^n \frac{i}{2^i} =\frac{2^{n+1}-n-2}{2^{n}} \,.$$
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