What is $\sum_{i = 0}^{n} (2^{ki})$?

Question

What is the result of $\sum\limits_{i = 0}^{n}(2^{ki})$

Is it $\frac{1-2^{kn}}{1-2^k}$?

Answer 1

$$\sum_{i=0}^{n}2^{ki}=\sum_{i=0}^{n}(2^k)^i=\frac{(2^k)^{n+1}-1}{2^k-1}=\frac{2^{kn+k}-1}{2^k-1}$$

Answer 2

Starting with \begin{align} \sum_{k=0}^{n} x^{k} &= \left( \sum_{k=0}^{n} + \sum_{k=n+1}^{\infty} - \sum_{k=n+1}^{\infty} \right) \, x^{k} \\ &= \sum_{k=0}^{\infty} x^{k} - \sum_{k=0}^{\infty} x^{k+n+1} \\ &= \frac{1 - x^{n+1}}{1-x} \end{align} then for $x = t^{m}$ it is given that $$\sum_{k=0}^{n} t^{m \, k} = \frac{1 - t^{m \, (n+1)}}{1 - t^{m}}.$$ Letting $t = 2$ provides the result in question.