# Thread: Proof sum of geometric series

1. ## Proof sum of geometric series

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 ....

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

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

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.

~~~~~

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.

~~~~~

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

try it

3. ## Re: Proof sum of geometric series

Originally Posted by romsek

try it

May I know you are referring to part 1 or part 2 of the question??

I just now solve part 2 using polynomial multiplication (see update original post).

Waiting to solve part 1.

Thank you....

4. ## Re: Proof sum of geometric series

Originally Posted by shakra
Waiting to solve part 1.
When $n=1$ we have $(1-x)(1+x^1)=(1-x^2)=1-x^{1+1}$ check n=1.

If $(1-x)(1+x+\cdots x^n)=1-x^{n+1}$ is known to be true then look at:

\begin{align*}(1-x)(1+x+\cdots+x^n+ x^{n+1})&=\underbrace {(1-x)(1+x+\cdots+x^n)}_{\text{Inductive step}}+(1-x)( x^{n+1}) \\&=(\underbrace {1-x^{n+1}}_{\text{result}})+(1-x)(x^{n+1})\\&=1-x^{n+2} \end{align*}\text{QED}

5. ## Re: Proof sum of geometric series

Thank you sir. for the solution actually method is the same provided..... but for sure interesting to have another comment.

In $p_{1}$ step you

Originally Posted by Plato
When $n=1$ we have $(1-x)(1+x^1)=(1-x^2)=1-x^{1+1}$ check n=1.
SOLUTION $p_{1}$ step

$(1-x)1 = 1-x^{1}$

I believe we have error in the solution right? should be $(1-x)(1+x)=(1-x^2)$ ???

By the way This solution from Introduction to Abstract Algebra W. Nicholson 2nd Edition p28 . Maybe they fix in latest (i can see it is now "4 edition" but my uni with only 2nd )

Originally Posted by Plato
When $n=1$ we have $(1-x)(1+x^1)=(1-x^2)=1-x^{1+1}$ check n=1.

If $(1-x)(1+x+\cdots x^n)=1-x^{n+1}$ is known to be true then look at:

\begin{align*}(1-x)(1+x+\cdots+x^n+ x^{n+1})&=\underbrace {(1-x)(1+x+\cdots+x^n)}_{\text{Inductive step}}+(1-x)( x^{n+1}) \\&=(\underbrace {1-x^{n+1}}_{\text{result}})+(1-x)(x^{n+1})\\&=1-x^{n+2} \end{align*}\text{QED}

6. ## Re: Proof sum of geometric series

Originally Posted by shakra
Thank you sir. for the solution actually method is the same provided..... but for sure interesting to have another comment. In $p_{1}$ step you
SOLUTION $p_{1}$ step
$(1-x)1 = 1-x^{1}$
I believe we have error in the solution right? should be $(1-x)(1+x)=(1-x^2)$ ???
By the way This solution from Introduction to Abstract Algebra W. Nicholson 2nd Edition p28 . Maybe they fix in latest (i can see it is now "4 edition" but my uni with only 2nd
@shakra, you say that you are located in London. I assume that is in the UK. But I must be mistaken. Having studied and taught in the UK, I know that no one as weak as you appear to be would have achieved a level upon which a pupil would be give this question.

You simply do not have the basic skills necessary to understand the solutions to these questions.
Over&out.

7. ## Re: Proof sum of geometric series

Originally Posted by Plato
@shakra, you say that you are located in London. I assume that is in the UK. But I must be mistaken. Having studied and taught in the UK, I know that no one as weak as you appear to be would have achieved a level upon which a pupil would be give this question.

You simply do not have the basic skills necessary to understand the solutions to these questions.
Over&out.
Wow. How can someone like become a moderator with such bad attitude? I must say that Any good forum should only encourage helping newbies. If you dont like it dont reply. Such insults can only be damaging. Really sad.

So..... Anyone else please answer this question or just explain me why I am so weak not to understand the answer.

I believe we have error in the solution right? should be $(1-x)(1+x)=(1-x^2)$ ???

8. ## Re: Proof sum of geometric series

Here's how I would have done it: $\displaystyle (1- x)(1+ x+ x^2+ \cdot\cdot\cdot+ x^n)= 1(1+ x+ x^2+ \cdot\cdot\cdot+ x^n)+ x(1+ x+ x^2+\cdot\cdot\cdot+ x^n)= (1+ x+ x^2+ \cdot\cdot\cdot+ x^n)- x(1+ x+ x^2+ \cdot\cdot\cdot+ x^n)= (1+ x+ x^2+ \cdot\cdot\cdot+ x^n)- (x+ x^2+ x^3+ \cdot\cdot\cdot+ x^{n+1})= 1- x^{n+1}$.

romsek may have been harsh but "Advanced Algebra by W. Nicholson" is intended for an advanced University course but the techniques necessary to do this problem, either direct multiplication and addition as I did (and was suggested earlier by romsek) or induction, are secondary school topics.

9. ## Re: Proof sum of geometric series

Originally Posted by HallsofIvy
romsek may have been harsh
romsek had nothing to do with this thread. I believe you mean Plato.

10. ## Re: Proof sum of geometric series

Originally Posted by romsek