# Thread: Geometric Progressions

1. ## Geometric Progressions

Hi all.

I'm trying to prove that

( 1 + x + x^2 + x^3 + ... + x^2k) ( 1 - x + x^2 - x^3 + ... + x^2k)

= 1 + x^2 + x^4 + ... + x^4k, where k is a positive integer and x is not equal to -1, 1.

This question was posted in the section on Geometric Progressions.

I don't know how to begin with this... is x^2k the last term? What does it mean, is it a power of 2?? For example x raised to some power of 2??

Do I use the summation formula S_n = a ( 1 - r^n) / (1 -r)??

Help will be appreciated.

Thanks.

2. Originally Posted by pollardrho06
Hi all.

I'm trying to prove that

( 1 + x + x^2 + x^3 + ... + x^2k) ( 1 - x + x^2 - x^3 + ... + x^2k)

= 1 + x^2 + x^4 + ... + x^4k, where k is a positive integer and x is not equal to -1, 1.

This question was posted in the section on Geometric Progressions.

I don't know how to begin with this... is x^2k the last term? What does it mean, is it a power of 2?? For example x raised to some power of 2??

Do I use the summation formula S_n = a ( 1 - r^n) / (1 -r)??

Help will be appreciated.

Thanks.
Yes, use the summation formula...you are summing n=2k+1 terms

$\displaystyle 1+x+x^2+.....x^{2k}$

the nth term is $\displaystyle ar^{n-1}\ \Rightarrow\ for\ x^{2k}\ \Rightarrow\ n=2k+1$

$\displaystyle a=1,\ r=x$

$\displaystyle S_n=\frac{1-x^{2k+1}}{1-x}$

$\displaystyle 1-x+x^2-x^3+...+x^{2k}$

$\displaystyle r=-x$

$\displaystyle S_n=\frac{1-(-x)^{2k+1}}{1-(-x)}$

2k+1 is odd as 2k is even, hence

$\displaystyle (-x)^{2k+1}=-\left(x^{2k+1}\right)$

$\displaystyle S_n=\frac{1+x^{2k+1}}{1+x}$

The product of these sums is

$\displaystyle \left(\frac{1-x^{2k+1}}{1-x}\right) \frac{1+x^{2k+1}}{1+x}=\frac{1-x^{4k+2}}{1-x^2}$

$\displaystyle 1+x^2+x^4+....+x^{4k}=1+(x)^2+\left(x^2\right)^2+. ...+\left(x^{2k}\right)^2$

$\displaystyle S_n=\frac{1-\left(x^2\right)^{2k+1}}{1-x^2}=\frac{1-x^{4k+2}}{1-x^2}$

3. Originally Posted by Archie Meade
Yes, use the summation formula...you are summing n=2k+1 terms

$\displaystyle 1+x+x^2+.....x^{2k}$

the nth term is $\displaystyle ar^{n-1}\ \Rightarrow\ for\ x^{2k}\ \Rightarrow\ n=2k+1$

$\displaystyle a=1,\ r=x$

$\displaystyle S_n=\frac{1-x^{2k+1}}{1-x}$

$\displaystyle 1-x+x^2-x^3+...+x^{2k}$

$\displaystyle r=-x$

$\displaystyle S_n=\frac{1-(-x)^{2k+1}}{1-(-x)}$

2k+1 is odd as 2k is even, hence

$\displaystyle (-x)^{2k+1}=-\left(x^{2k+1}\right)$

$\displaystyle S_n=\frac{1+x^{2k+1}}{1+x}$

The product of these sums is

$\displaystyle \left(\frac{1-x^{2k+1}}{1-x}\right) \frac{1+x^{2k+1}}{1+x}=\frac{1-x^{4k+2}}{1-x^2}$

$\displaystyle 1+x^2+x^4+....+x^{4k}=1+(x)^2+\left(x^2\right)^2+. ...+\left(x^{2k}\right)^2$

$\displaystyle S_n=\frac{1-\left(x^2\right)^{2k+1}}{1-x^2}=\frac{1-x^{4k+2}}{1-x^2}$
Gr8, thanks!!