<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] Notice that
$$\sum_{n=1}^{100}x^n=\frac{x-x^{101}}{1-x}$$
which is just a geometric series.  Differentiate once to get
$$\sum_{n=1}^{100}nx^{n-1}=\frac d{dx}\frac{x-x^{101}}{1-x}$$
Multiply both sides by $x$ and set $x=-\frac12$ to get
$$\frac1{2^1}-\frac2{2^2}+\dots-\frac{100}{2^{100}}=-\frac12\left(\frac d{dx}\bigg|_{x=-\frac12}\frac{x-x^{101}}{1-x}\right)$$
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