I try to solve the following sum:
$\sum_{k=0}^{n}\frac{2}{3^k} = \frac{2}{1} + \frac{2}{3} + \frac{2}{9} + \frac{2}{81} + ...$
The recurring factor of all numbers is $\frac{1}{3}$, for example $\frac{2}{1} \times \frac{1}{3} = \frac{2}{3}$ and $\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$.
According to the known formula for geometric sums:
$a + aq + aq^2 + ... + aq^n = \frac{a(q^{n+1} - 1)}{q - 1}$
where $q$ is the factor and $n+1$ is the number of terms. Therefore:
$\sum_{k=0}^{n}\frac{2}{3^k} = \frac{\frac{2}{1}((\frac{1}{3})^{n+1} - 1)}{\frac{1}{3} - 1} = \frac{2((\frac{1}{3})^{n+1} - 1)}{-\frac{2}{3}} = $
$= \frac{6((\frac{1}{3})^{n+1} - 1)}{-2} = -3((\frac{1}{3})^{n+1} - 1) =$
$= -3((\frac{1}{3})^{n} \times (\frac{1}{3})^{1} - 1) = -3(\frac{1}{3^{n}} \times \frac{1}{3} - 1) =$
$= \frac{-3}{3^{n}} \times \frac{-3}{3} + 3 = \frac{-3}{3^{n}} \times (-1) + 3 =$
$= \frac{3}{3^{n}} + 3$
This is clearly not the answer. For example when $n = 0$, this equation shows a term of $4$, which doesn't exist.
Where am I going wrong?