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Value of $\sum\limits_n x^n$

I don't know the technical language for what I'm asking, so the title might be a little misleading, but hopefully I can convey my purpose to you just as well without.

Essentially I'm thinking of this: the series $4^n + 4^{n-1} \cdots 4^{n-n}$.

I suppose this is the summation of the series $4^n$ from $n$ to 0.

But is there any way to express this as a pure equation, not as a summation of a series?

If so, how do you figure out how to convert it?

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marked as duplicate by Marvis, Eugene, Ross Millikan, Zev Chonoles Jun 26 '12 at 4:36

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You may find the post math.stackexchange.com/questions/29023/value-of-sum-limits-n-xn useful. –  user17762 Jun 26 '12 at 4:34

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up vote 2 down vote accepted

In general, for $x\neq 1$ it is true that $$\sum_{k=0}^nx^k=1+x+\cdots+x^n=\frac{x^{n+1}-1}{x-1}.$$ So, in your case in particular, we have that $$\sum_{k=0}^n4^{n-k}=4^n+\cdots+4+1=1+4+\cdots+4^n=\sum_{k=0}^n4^k=\frac{4^{n+1}-1}{3}.$$ Alternatively, one could pull out a factor of $4^n$ from all terms, and compute $$\sum_{k=0}^n4^{n-k}=4^n\sum_{k=0}^n(\tfrac{1}{4})^k=4^n\cdot\frac{(\frac{1}{4})^{n+1}-1}{(\frac{1}{4})-1}=4^n\cdot\frac{\frac{4^{n+1}-1}{4^{n+1}}}{\frac{3}{4}}=4^{n+1}\cdot\frac{\frac{4^{n+1}-1}{4^{n+1}}}{3}=\frac{4^{n+1}-1}{3}.$$

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