Hello guys

I got a problem in the question of "proof of Sum of geometric series" $1+x+...+x^n$ and infact uses the polynomial multiplication $(1-x)(1+x+x^2+...+x^n) = 1-x^{n+1}$ Anyway, see you can help me ....

$(1+x)(1+x+...+x^{n-1})=1-x^n$

If x is any real number, show that for $n\geq1$

SOLUTION

Let $p_{n}$ the given statement. Then $p_{1}$ is $(1-x)1=1-x^1$ which is true

If we assume $p_{k}$ true for some $k\geq1$ then

$(1-x)(1+x+...+x^{k-1}+x^k)=(1-x)(1+x+...+x^{k-1})+(1-x)x^k$

$= (1-x^k)+(1-x)x^k$

$= 1-x^{k+1}$

This proves that $p_{k+1}$ is true and completes the induction.

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QUESTION

First this line

Let $p_{n}$ the given statement. Then $p_{1}$ is $(1-x)1=1-x^1$ which is true

If we sub in $n=1$ to $p_{n}$ we get

$p_{n}=(1+x)(1+x+...+x^{n-1})$

$(1+x)(1+x+...+x^{1-1})$

$=(1+x)(x^0)$

$=1(1+x)\neq (1-x)1$

so for sure I dont understand this line in the solution.

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EDIT NOW already solved this "part 2"

use (from "polynomial multiplication"): $(1-x)(1+x+x^2+...+x^n) = 1-x^{n+1}$

so: $(1-x)(1+x+...+x^{k-1}) = (1-x^k)$

Second......

How to get from line $(1-x)(1+x+...+x^{k-1})+(1-x)x^k$ to next line $= (1-x^k)+(1-x)x^k$

Dont understand what happened there..

Hope you guys give a little help to me