Geometric Progressions

**pollardrho06** · May 20th 2010, 10:56 AM

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.

**Archie Meade** · May 20th 2010, 01:20 PM

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

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

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

$a=1,\ r=x$

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

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

$r=-x$

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

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

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

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

The product of these sums is

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

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

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

**pollardrho06** · May 20th 2010, 01:28 PM

Originally Posted by Archie Meade

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

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

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

$a=1,\ r=x$

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

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

$r=-x$

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

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

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

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

The product of these sums is

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

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

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

Gr8, thanks!!

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