If $n \in \mathbb N$, find $\sum(-1)^r \binom{n}{r}\left(\frac {1} {2^r}+\frac {3^r} {2^{2r}}+\frac {7^r} {2^{3r}} + \cdots \text{m terms}\right)$

Question

If $n \in \mathbb N$, find$$\sum(-1)^r \binom{n}{r}\left(\frac {1} {2^r}+\frac {3^r} {2^{2r}}+\frac {7^r} {2^{3r}} + \frac {15^r}{2^{4r}} + \cdots \text{upto m terms}\right)$$

Answer 1

We can rewrite the sum as:

$$\sum \binom {n} {r} \left(- \frac {1} {2}\right)^r + \sum \binom {n} {r} \left(- \frac {3} {2^2}\right)^r + \sum \binom {n} {r} \left(- \frac {7} {2^3}\right)^r + \cdots \text {upto m terms} $$

$$= \left( 1- \frac{1} {2}\right)^n +\left( 1- \frac{3} {2^2}\right)^n + \left( 1- \frac{7} {2^3}\right)^n + \cdots \text{upto m terms}$$

$$= \frac{1} {2^n} +\frac{1} {(2^n)^2} +\frac{1} {(2^n) ^3} + \cdots \text {upto m terms} $$

Solving for sum of geometric series upto $m$ terms, we get:

$$\frac {2^{mn} - 1} {(2^n - 1)2^{mn}}$$