Evaluate the following expression:

Question

$$\sum_{r=0}^n (-1)^r {n\choose r} \left[\left(\frac{1}{2}\right)^r + \left(\frac{3}{2^2}\right)^r + \left(\frac{7}{2^3}\right)^r + \left(\frac{15}{2^4}\right)^r+ \cdots m\text{-terms}\right]$$

Please provide me with a solution. A hint would do too. Thank you.

Answer 1

This summation can be broken up and written like:

\begin{align} & \sum_{r=0}^n (-1)^{r}{n\choose r} \left[\left(\frac{1}{2}\right)^r + \left(\frac{3}{2^{2}}\right)^r + \left(\frac{7}{2^3}\right)^r + \left(\frac{15}{2^4}\right)^r+ \cdots\ m\text{-terms}\right] \\[10pt] = {} & \sum_{r=0}^n (-1)^r {n\choose r} \left(\frac{1}{2}\right)^r+\sum_{r=0}^n (-1)^r {n\choose r} \left(\frac{3}{4}\right)^r + \sum_{r=0}^n (-1)^{r}{n\choose r} \left(\frac{7}{8}\right)^r+\cdots \\[10pt] = & {} \left(1-\frac{1}{2}\right)^n+\left(1-\frac{3}{4}\right)^n+\left(1-\frac{7}{8}\right)^n+\cdots \end{align}

(Using the fact that $\sum_{r=0}^n (-1)^r {n\choose r}x^{r}=(1-x)^n$)

And you're almost done. (Look for a Geometric progression)