Geometric Progressions

Feb 2010
36
2
New Jersey, USA
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.
 
Dec 2009
3,120
1,342
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

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


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

\(\displaystyle S_n=\frac{1-\left(x^2\right)^{2k+1}}{1-x^2}=\frac{1-x^{4k+2}}{1-x^2}\)
 
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Reactions: pollardrho06
Feb 2010
36
2
New Jersey, USA
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!!