# Getting wrong result calculating sums

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?

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In the previous to last line of displayed equations, you add erroneously a minus sign. –  Did Aug 24 '12 at 9:12
Part of the problem may be that you were carrying too many minus signs around, they are heavy. For $q\lt 1$, I would probably almost automatically use the formula $a\frac{1-q^{n+1}}{1-q}$, or after using your formula switch signs (multiply top and bottom by $-1$). –  André Nicolas Aug 24 '12 at 13:05

$$-3\left(\frac{1}{3^{n}} \times \frac{1}{3} - 1\right) \neq \frac{-3}{3^{n}} \times \frac{-3}{3} + 3$$

There is no a distributive property with $a×(b×c) = (a×b)×(a×c)$! Be careful!

$$-3\left(\frac{1}{3^{n}} \times \frac{1}{3} - 1\right) = \frac{-3}{3^{n}} \times \frac{1}{3} + 3$$

$$= \frac{-1}{3^{n}} + 3$$

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Then you have miscalculated the (existing incorrect) formula for $n=0$ - instead of 4 you should get 6.