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

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

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 $.

 

[blue_raver22]

Hello! This is a great question and I would be happy to help you with the proof.

To start, let's expand the right side of the equation using the formula for the sum of a geometric series:

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

Next, we can see that the last term on the right side, z^(n+1), is not part of the original equation. So, we can subtract it from both sides:

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

Now, let's factor out z^(n+1) from the left side:

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

Using the formula for the sum of a geometric series again, we can simplify the right side:

z^(n+1) (1/(1-z) - 1) = (1-z^(n+1))/(1-z)

Finally, we can see that the left side is equal to the right side, so we have proven the original equation:

(1-z^(n+1))/(1-z) = (1-z^(n+1))/(1-z)

I hope this helps! Let me know if you have any further questions. Keep up the good work!