$$\sum\limits_{n = 10}^{40} \left(\frac{1}{8}\right)^n$$

I do not know the steps to finding the sum of this problem.

I know the general formula is $$ = a_1\frac{1 - r^n}{1 - r}.$$

Wolfram alpha link

  • $\begingroup$ Please add the left side of the equation for the "general formula". $\endgroup$ – dfeuer Aug 8 '13 at 5:29

You could do this two ways: pull out a factor of $\left(\frac{1}{8}\right)^{10}$ from the sum $$\sum_{n=10}^{40}\left(\frac{1}{8}\right)^n$$ to make $$\left(\frac{1}{8}\right)^{10}\cdot\sum_{n=0}^{30}\left(\frac{1}{8}\right)^n$$ (and this new sum you should know how to do), or you can write the sum $$\sum_{n=10}^{40}\left(\frac{1}{8}\right)^n$$ as a difference $$\sum_{n=0}^{40}\left(\frac{1}{8}\right)^n\;\;-\;\;\sum_{n=0}^{9}\left(\frac{1}{8}\right)^n$$ (and again, each of these two sums you should know how to do).


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