Summation over exponent $\sum_{i=0}^k 4^i= \frac{4^{k+1}-1}3$ Why does $$\sum_{i=0}^k 4^i= \frac{4^{k+1}-1}{3}$$where does that 3 comes from?
Ok, from your answers I looked it up on Wikipedia Geometric Progression, but to derive the formula it says to multiply by $(1-r)$ not $(r-1)$ why is this case different?

Recall that $$\sum\limits_{i = 0}^k {a{r^i} = a + ar + a{r^2} +  \ldots  + a{r^k} = a\frac{{{r^{k + 1}} - 1}}{{r - 1}}}$$
(see my answer to this question if you wish to prove the above result).
