Evaluate: $\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^2}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$

Question

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?

Answer 1

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:

$$ \sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n} = \color{red}{\frac{1}{9}\left(2-\frac{302}{2^{100}}\right)}.$$

Answer 2

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

$$\frac1{2^1}-\frac2{2^2}+\dots-\frac{100}{2^{100}}=-\frac12\left(\frac d{dx}\bigg|_{x=-\frac12}\frac{x-x^{101}}{1-x}\right)$$

Answer 3

$$I=\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$ thus $$2I=1-\frac{2}{2^1}+\frac{3}{2^2}-\frac{4}{2^3}+\ldots-\frac{100}{2^{99}}$$ as a result $$2I+I=1-\left(\frac{2}{2^1}-\frac{1}{2^1}\right)+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)-\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\cdots-\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)-\frac{100}{2^{100}}$$ therefore $$3I=\underbrace{\left(1-\frac 12+\frac 14-\frac 18+\cdots-\frac{1}{2^{99}}\right)}_{\frac{2^{101}-2}{3\times 2^{100}}}-\frac{100}{2^{100}}=\frac{2^{101}-302}{3\times 2^{100}}$$