So 1−x21=x2+x4+x6...?divide
[MATH]\dfrac{1}{1-x^2} = 1 \text{ with remainder $x^2$}\\ \dfrac{x^2}{1-x^2} = x^2 \text{ with remainder $x^4$}\\ \dfrac{x^4}{1-x^2} = x^4 \text{ with remainder $x^6$}\\ \text{etc}[/MATH]
looks like you'll end up with [MATH]\dfrac{1}{1- x^2} = \sum \limits_{k=1}^\infty x^{2k}[/MATH]
Check your answer using Taylor's expansion.So 1−x21=x2+x4+x6...?
NO!So 1−x21=x2+x4+x6...?
I didNO!
You lost the leading 1 of the series!! I was hoping you would catch it by now.......