Can you help me out with this? I always get in a pickle. How can one show that $1+r+r^2+...+r^n=\dfrac{1-r^{n+1}}{1-r}$ for $n=0,1,2,3... $and $r \neq 1$.
Proving that $\sum_{k=0}^{n} r^k=\dfrac{1-r^{n+1}}{1-r}$ when $r \neq 1$.
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algebra-precalculus
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3Could I? Yes. Will I? Nope. Not until you fix your title, format your post, or *at least* put a little effort/context into this. MSE is *not* a site to throw your homework problems on and expect us to do the work for you – 2017-02-10
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0Prove first that $(1-r)(1+r)=1-r^2\,$. Then $(1-r)(1+r+r^2)=1-r^3\,$. Then confirm the pattern. – 2017-02-10
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1Possible duplicate of [Geometric representation of the sum of a Geometric Series](http://math.stackexchange.com/questions/371612/geometric-representation-of-the-sum-of-a-geometric-series) – 2017-02-10
2 Answers
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This is more or less the formula for the partial sums of a geometric series.
One can prove it in the following way:
Set $A=1+...+r^n$. Then $A \cdot r=r+...+r^{n+1}$. Subtracting the two equality yields:
$$A-Ar=1-r^{n+1}.$$
Factor both sides by $A$ and divide both sides by $1-r$ to finish. Recall that $A$ is the actual sum, so we can replace the equality
$A=\frac{1-r^{n+1}}{1-r}$ with $1+...+r^n=\frac{1-r^{n+1}}{1-r}$ by substitution.
It should be clear that $r \neq 1$, since otherwise the division would be invalid.
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You can prove the assertion by induction. The case $n=0$ is clear and in the step from $n$ to $n+1$ you have to show that
$$\frac{1-r^{n+1}}{1-r}+r^{n+1}=\frac{1-r^{n+2}}{1-r}.$$
Try it !