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.

  • 1
    $\begingroup$ Would writing it as $(1/6)^n$ unstub you? $\endgroup$ – David Mitra Mar 8 '15 at 17:32
  • $\begingroup$ Its that all the series I have done it has been i and not n. $\endgroup$ – user2079139 Mar 8 '15 at 17:34
  • $\begingroup$ @DavidMitra how is it a series if its to the power of n. Its not incrementing? $\endgroup$ – user2079139 Mar 8 '15 at 17:35
  • $\begingroup$ 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. $\endgroup$ – David Mitra Mar 8 '15 at 17:38
  • $\begingroup$ @DavidMitra nope its 1/6^n $\endgroup$ – 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.
Then the sum transforms into
$$\sum_{i=0}^{n}k=(n+1)k$$ 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 $$(n+1)\left (\frac{1}{6}\right )^n$$


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