What are the coefficients in the expansion of the following expression at x=0: (1 + x)^(-2) and how are they calculated?

Thank you for help.

 Oct 16, 2018
edited by Guest  Oct 16, 2018

The most straightforward way to do this is to just do the division.

It's hard to show the process of long division on here.




and just set this up as a usual long division, if you go through it a few steps you'll see it's


\(\dfrac{1}{1+2x+x^2} = 1 - 2x + 3x^2 - 4x^3 + 5x^4 \dots + (-1)^k (k+1)x^k,~\forall k \geq 0\)


hrm.. there might be another way to approach it that's pretty straightforward


\(\sum \limits_{k=0}^\infty ~x^k = \dfrac{1}{1-x}\)




\(\dfrac{1}{1-(-x)} = \sum \limits_{k=0}^\infty ~(-1)^k x^k\)


\(\dfrac{d}{dx}~\dfrac{1}{1+x} = - \dfrac{1}{(1+x)^2}\)


\(\dfrac{1}{(1+x)^2} = -\dfrac{d}{dx} \left(\sum \limits_{k=0}^\infty~(-1)^k x^k\right) = \\ -\left(\sum \limits_{k=1}^\infty~(-1)^k k x^{k-1}\right) = \sum \limits_{k=0}^\infty~(-1)^k(k+1)x^k\)

 Oct 16, 2018

Thank you very much Rom. The 2nd method is very easy to follow. Thanks again.

 Oct 16, 2018

14 Online Users


New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.