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Why is $5 + 5z + 5z^2 + ... + 5z^{11} = \frac{(5z^{12} - 5)}{(z - 1)}$ ? I don't understand how you can rewrite it to that. Z is in this case a complex number: (for example: $z = 0,8(0,5 + 0,5i\sqrt{3}) = 0,4 + 0,4i\sqrt{3}$). Is it also possible to write the $5 + 5z + 5z^2 + ... + 5z^{11}$ with a summation (sigma) sign?

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    $\begingroup$ It's a geometric sum. To verify, multiply the left and right by $z-1$, things will cancel. $\endgroup$ – Tob Ernack Mar 14 '18 at 15:35
  • $\begingroup$ It's the sum of a geometric progression. $\endgroup$ – Angina Seng Mar 14 '18 at 15:35
  • $\begingroup$ $$5\left(\frac{z^{12}-1}{z-1}\right)$$ is found via the formula for a geomteric series. see mathworld.wolfram.com/GeometricSeries.html $\endgroup$ – Hyperkähler Mar 14 '18 at 15:38
  • $\begingroup$ Did you try seeing what $(z-1)(5 + 5z + 5z^2 + ... +5z^{11})$. $\endgroup$ – fleablood Mar 14 '18 at 15:42
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For $z\neq 1$

$$5 + 5z + 5z^2 + … + 5z^{11} = \frac{(5z^{12} - 5)}{(z - 1)}\iff \\\iff (z-1)(5 + 5z + 5z^2 + … + 5z^{11}) = (5z^{12} - 5)$$

which is true by direct inspection, indeed

$$z\cdot (5 + 5z + 5z^2 + … + 5z^{11} ) = 5z + 5z^2 + … + 5z^{12} $$

$$-1\cdot (5 + 5z + 5z^2 + … + 5z^{11}) = -5 - 5z - 5z^2 + … - 5z^{11} $$

then sum up.

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The proof is only true for $x \ne 1$:

$$\frac{5z^{12}-5}{z-1}=\frac{5(z^{12}-1)}{(z-1)}=\frac{5(z-1)(1+z++z^2+z^3+...+z^{10}+z^{11})}{z-1}=5(1+z+z^2+z^3+...+z^{10}+z^{11})$$

There is an equality you need to prove for the expression above, which is $a^{n+1}-b^{n+1}=(a-b)(a^{n}b^{0}+a^{n-1}b^{1}+a^{n-2}b^2+...+a^{1}b^{n-1}+a^{0}b^{n})$, this question will help you prove it and complete the proof.

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An infinite geometric series $a+ar+ar^2+ar^3+\cdots$, where $a$ and $r$ are fixed, has a tidy closed-form formula:

$$ a+ar+ar^2+ar^3 = \frac{a}{1-r} $$

Now, your series isn't infinite, but note that it can be expressed as the difference between two series that are infinite:

$$ 5+5z+5z^2+5z^3+\cdots+5z^11 = (5+5z+5z^2+5z^3+\cdots)-(5z^{12}+5z^{13}+5z^{14}+\cdots) $$

Provided you stipulate that $z \not= 1$, if you apply the above formula to the two infinite series above, subtract, and do a little algebraic clean-up, you should obtain the desired expression.

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Let $$S=\sum_{i=0}^n az^i, \quad z\ne1.$$

Then $$S\cdot z=\sum_{i=0}^n az^{i+1}= S+az^{n+1}-a \Rightarrow S(z-1)=az^{n+1}-a \Rightarrow S=\frac{az^{n+1}-a}{z-1}.$$

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That is because $$1+z+z^2+\dots+z^n=\frac{z^{n+1}-1}{z-1}$$ from a high-school formula for factorisation: $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b+\dots+ab^{n-2}+b^{n-1}).$$

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Arnaud D. Mar 14 '18 at 16:05
  • $\begingroup$ @ArnaudD. Do you think the O.P. doesn't understand ellipsis? $\endgroup$ – Bernard Mar 14 '18 at 16:13
  • $\begingroup$ I think that if OP doesn't know how to compute $5+5z+\dots +5z^{11}$, they probably have forgotten (or never learned) the formula for the sum of a geometric series. The added formula is already more helpful. $\endgroup$ – Arnaud D. Mar 14 '18 at 16:21
  • $\begingroup$ I'm going to add a compensatory upvote here. I just think the answer could be reworded—perhaps as an observation that if you multiply $5+5z+\cdots+5z^{11}$ by $z-1$, a telescoping occurs that yields only the two "endpoints" of the series. $\endgroup$ – Brian Tung Mar 14 '18 at 17:03

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