Prove that the following identity is valid for all $z$ with $z\neq1$:
$$1+z+z^2+...+z^n=\frac{1-z^{n+1}}{1-z}$$
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1This is a standard fact about geometric series that is proved in hundreds of books, the wikipedia article about geometric series, and on this very site many times. Have you done any research about this problem? – Aug 23 '16 at 19:35
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Yes, I have. I was just curious on how to go about it. So, would standard Mathematical Induction be all? It seems too simple for me. I can post my solution shortly! – Analysis15 Aug 23 '16 at 19:37
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Yes, that is one of a very large number of ways to prove it. (And had you included that in your original question, it probably wouldn't have immediately been downvoted) – Aug 23 '16 at 19:37
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The result is immediate by synthetic division. – Aug 23 '16 at 19:48
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Key words : sum finite geometric series – Jean Marie Aug 23 '16 at 19:58
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Duplicate of http://math.stackexchange.com/questions/11703/proof-of-the-formula-1xx2x3-cdots-xn-fracxn1-1x-1?noredirect=1&lq=1 – Watson Aug 23 '16 at 20:01
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\begin{align} S=1+&z+z^2+z^3+\cdots+z^n\\ zS=\quad\quad&z+z^2+z^3+\cdots+z^n+z^{n+1}\\ \\ S-zS=1-z^{n+1}\\ S=\frac{1-z^{n+1}}{1-z} \end{align}
Kay K.
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1Note that a far more explained version of this same answer is given in the top-voted answer of the faq version of this post. – Aug 23 '16 at 19:50
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Multiply out $(1-z)(1+z+z^2+\dots+z^n)$.
You get $(1+z+z^2+\dots+z^n)-(z+z^2+z^3+\dots+z^{n+1})=1-z^{n+1}$.
pi66
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By induction:
With $n=0$, $$1=\frac{1-z^1}{1-z}.$$
Then
$$S_n+z^{n+1}=\frac{1-z^{n+1}}{1-z}+z^{n+1}=\frac{1-z^{n+2}}{1-z}=S_{n+1}.$$
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$$S_{n+1}=\frac{1-z^{n+1}}{1-z}=\frac{1-z+z-z^{n+1}}{1-z}=1+zS_{n},$$ with $S_0=1$.
Then the iterates are
$$S_0=1,S_1=1+z,S_2=1+z+z^2,S_3=1+z+z^2+z^3,\cdots$$