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?

The other answers are correct, this is the finite geometric series for $r=4$. You asked why this is so, and in fact, this is not too difficult to derive. Let $S$ be the sum we want to evaluate, which means
$S=1+r+r^2+r^3+\ldots+r^n$  
If we multiply by $r$,
$rS=r+r^2+r^3+\ldots+r^n+r^{n+1}$
Now subtract
$rS-S = S(r-1) = r^{n+1}-1$
$\Longrightarrow S= \dfrac{r^{n+1}-1}{r-1}$.
So for your series, evaluating this for $r=4$ gives $S=\dfrac{4^{n+1}-1}{4-1}$. Thus the $3$ comes from the $r-1$ term in the denominator.
Note: In the final step we are dividing by $r-1$, so this formula is no longer valid if $r=1$, since we cannot divide by zero.
Comment response: In the way it's done on Wikipedia, the sum is $S=\dfrac{1-r^{n+1}}{1-r}$, and actually this is equivalent to what I wrote. Note that $(1-r^{n+1})=-(r^{n+1}-1)$ and the same goes for the denominator; $1-r = -(r-1)$. So since the sign of the terms change on both top and bottom, the fraction remains the same. Thus both expressions are equivalent.
