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- #1

- Thread starter shen07
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- Thread starter
- #1

- Feb 13, 2012

- 1,704

It is immediate to verify that is...Hello Guys once again need your help for a proof.

Prove

1+z+Z^2+.....+z^n=(1-z^(n+1))/(1-z)

$\displaystyle z^{n+1} - 1 = (z-1)\ (1 + z + z^{2} + ... + z^{n})\ (1)$

Kind regards

$\chi$ $\sigma$

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- #3

- Jan 17, 2013

- 1,667

You can use a proof by induction , even though the way chisigma suggested suffices

\(\displaystyle \tag{1}1+z+z^2+ \cdots +z^n = \frac{z^{n+1}-1}{z-1}\)

**Base case **

is \(\displaystyle n = 0 \) we get $1$

__Inductive step __

Assume that (1) is correct and want to prove

\(\displaystyle \tag{2}1+z+z^2+ \cdots +z^{n+1} = \frac{z^{n+2}-1}{z-1}\)

From (1)

\(\displaystyle 1+z+z^2+ \cdots +z^n+z^{n+1}= \frac{z^{n+1}-1}{z-1}+z^{n+1}= \frac{z^{n+2}-1}{z-1}\)

Hence (2) is satisfied which completes the proof $\square $.

\(\displaystyle \tag{1}1+z+z^2+ \cdots +z^n = \frac{z^{n+1}-1}{z-1}\)

is \(\displaystyle n = 0 \) we get $1$

Assume that (1) is correct and want to prove

\(\displaystyle \tag{2}1+z+z^2+ \cdots +z^{n+1} = \frac{z^{n+2}-1}{z-1}\)

From (1)

\(\displaystyle 1+z+z^2+ \cdots +z^n+z^{n+1}= \frac{z^{n+1}-1}{z-1}+z^{n+1}= \frac{z^{n+2}-1}{z-1}\)

Hence (2) is satisfied which completes the proof $\square $.

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