<s>[INST] Evaluate: $\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^2}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$ The problem is to evaluate the following sum:
$$\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$
My approach was to find the common denominator ($2^{100}$), then the series becomes:
$$ \frac{1\cdot 2^{99}-2\cdot 2^{98}+3\cdot 2^{97}-4\cdot 2^{96}+\ldots -100\cdot 2^0}{2^{100}}$$
And split it this way:
$$ \frac{(1\cdot 2^{99}+3\cdot 2^{97}+\ldots+99\cdot 2^{1})-(2\cdot 2^{98}+4\cdot 2^{96}+\ldots+100\cdot 2^{0})}{2^{100}}$$
But it seems that cacelling out the terms is not a promising approach. Any hints? 
 [/INST] Try to multiply the given sum by $\frac{3}{2}=1+\frac{1}{2}$:
$$\begin{eqnarray*}\left(1+\frac{1}{2}\right)\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n}&=&\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n}+\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^{n+1}}\\&=&\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n}+\sum_{n=2}^{101}\frac{(-1)^{n}(n-1)}{2^{n}}\\&=&\frac{1}{2}-\frac{100}{2^{101}}-\sum_{n=2}^{100}\frac{(-1)^n}{2^n}\end{eqnarray*}$$
then perform the same manipulation with $\sum_{n=2}^{100}\frac{(-1)^n}{2^n} = \frac{1}{6}\left(1+\frac{1}{2^{99}}\right)$. That leaves you with:
$$ \sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n} = \color{red}{\frac{1}{9}\left(2-\frac{302}{2^{100}}\right)}.$$
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