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How can this be summed: $$\sum_{k=\frac{m}2}^{m-1}x^k$$

thank you!

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Factor $x^{m/2}$ out, use geometric sum formula. – anon Jul 11 '12 at 23:24

Just subtract one geometric series from another:


(I’m assuming from your title that you already know how to deal with these sums.)

Added: More generally, $$\sum_{k=m}^nx^k=x^m\sum_{k=0}^{n-m}x^k=x^m\left(\frac{x^{n-m+1}-1}{x-1}\right)=\frac{x^{n+1}-x^m}{x-1}=\frac{x^m-x^{n+1}}{1-x}$$ is the general formula for such sums.

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This is a partial geometric series. I am assuming that $m$ is even since you have $m/2$ as a limit.

Using the formula for the sum of a geometric series, we get

$$ \begin{align} \sum_{k=m/2}^{m-1}x^k &=x^{m/2}\sum_{k=0}^{m/2-1}x^k\\ &=x^{m/2}\frac{x^{m/2}-1}{x-1}\\ &=\frac{x^m-x^{m/2}}{x-1} \end{align} $$

Different Methods, Same Result:

We can compute $\displaystyle\sum_{k=m}^{n-1}ar^k$ in two ways:

  1. subtracting two geometric series: $$ \begin{align} \sum_{k=m}^{n-1}ar^k &=\sum_{k=0}^{n-1}ar^k-\sum_{k=0}^{m-1}ar^k\\ &=a\frac{r^n-1}{r-1}-a\frac{r^m-1}{r-1}\\ &=a\frac{r^n-r^m}{r-1}\tag{1} \end{align} $$

  2. Pulling out the starting power of $r$: $$ \begin{align} \sum_{k=m}^{n-1}ar^k &=r^m\sum_{k=0}^{n-m-1}ar^k\\ &=r^ma\frac{r^{n-m}-1}{r-1}\\ &=a\frac{r^n-r^m}{r-1}\tag{2} \end{align} $$

Both $(1)$ and $(2)$ yield the same result.

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