# Sum of A Finite Series $\sum_{i=0}^n \frac1{6^n}$

I am trying to solve the sum of this finite series: $$\sum_{i=0}^n \frac{1}{6^n}.$$ I am having problems where to start, as it is completely different to the other ones I have done.

Here is the Wolframalpha link: http://www.wolframalpha.com/input/?i=sum+1%2F6^n%2C+i%3D0+to+n

The $\dfrac{1}{6^n}$ just stubs me.

Thanks for the help.

• Would writing it as $(1/6)^n$ unstub you? – David Mitra Mar 8 '15 at 17:32
• Its that all the series I have done it has been i and not n. – user2079139 Mar 8 '15 at 17:34
• @DavidMitra how is it a series if its to the power of n. Its not incrementing? – user2079139 Mar 8 '15 at 17:35
• Oh, I missed that. As written, it's $(n+1)\cdot(1/6)^n$ (you're summing $n+1$ terms, all of which are $(1/6)^n$). But it may be a typo, and $(1/6)^i$ was meant. – David Mitra Mar 8 '15 at 17:38
• @DavidMitra nope its 1/6^n – user2079139 Mar 8 '15 at 17:40

Let $\left (\frac{1}{6}\right )^n=k$ for simplicity's sake ($k$ is a constant). It is possible in this case because there is no variable term in the summation. We can do this only when the terms are constant throughout the summation.
$$\sum_{i=0}^{n}k=(n+1)k$$ This is because $i$ starts from $0$ to $n$, and there are $(n+1)$ integers in between. $k$ just gets added $n+1$ times. Now, we just substitute the value of $k$ and get the answer as $$(n+1)\left (\frac{1}{6}\right )^n$$