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

[shen07]

Hello Guys once again need your help for a proof.

Prove

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

 

[chisigma]

Hello Guys once again need your help for a proof.

Prove

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

It is immediate to verify that is... $\displaystyle z^{n+1} - 1 = (z-1)\ (1 + z + z^{2} + ... + z^{n})\ (1)$

Kind regards

$\chi$ $\sigma$

 

[shen07]

yeah but how do we go about proving it..??

 

[alyafey22]

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

$$\tag{1}1+z+z^2+ \cdots +z^n = \frac{z^{n+1}-1}{z-1}$$

Base case

is $$n = 0 $$ we get $1$

Inductive step

Assume that (1) is correct and want to prove

$$\tag{2}1+z+z^2+ \cdots +z^{n+1} = \frac{z^{n+2}-1}{z-1}$$

From (1)

$$1+z+z^2+ \cdots +z^n+z^{n+1}= \frac{z^{n+1}-1}{z-1}+z^{n+1}= \frac{z^{n+2}-1}{z-1}$$

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