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?

Here is an easy mnemonic.  If you have a geometric sum, then
$$\sum {\rm geometric} = {{\rm first} - {\rm last}\over {1 - {\rm common \ Ratio}}}.$$
In this case, first is the first term, blast is the one beyond the last, and commonRatio is the common ratio of the terms.  If the sum is finite and ${\rm commonRatio} > 1$, reverse the subtractions in the numerator and denominator for greater prettiness.
