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Taylor series gives $$\frac 1 {(1+x)^2}=1-2x+3x^2-4x^3+\cdots$$

is there a nice expression for $1-4x+9x^2-16x^3 + \cdots ?$

It would be helpful for a problem I am trying to solve.

user85798
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    I voted to close as a duplicate, but it's not an exact duplicate: the original question has $\sum_{n=0}^{\infty} n^2x^n$, whereas this question asks for $\sum_{n=0}^{\infty} n^2(-x)^n$. – TonyK Nov 26 '15 at 13:14
  • @r9m Signs alternate in this question, but substituting $-x$ for $x$ in the link does the trick. The question isn't quite the same, though, and an extra step is required. – Mark Bennet Nov 26 '15 at 13:14
  • Yeah thanks guys, im fine now – user85798 Nov 26 '15 at 13:18

5 Answers5

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Here is a way - I'll leave you to do the detailed calculations.

Start with $f(x)=-1+x-x^2+x^3-\dots=-\cfrac 1{1+x}$

Then $f'(x)=1-2x+3x^2-4x^3+\dots$

And $xf'(x)=x-2x^2+3x^3-4x^4+\dots$

So that $\left(xf'(x)\right)'=1-4x+9x^2+-\dots$

I hope the method proves helpful

Mark Bennet
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For $|x| < 1$, we have $$\begin{align} \sum_{n=0}^\infty (n+1)^2 (-x)^n &= \sum_{n=0}^\infty \left(x\frac{d}{dx} + 1 \right)^2 (-x)^n = \left(x\frac{d}{dx} + 1 \right)^2 \sum_{n=0}^\infty (-x)^n\\ &= \left(x\frac{d}{dx} + 1 \right)^2 \frac{1}{1+x} = \frac{d}{dx}\left[ x \frac{d}{dx}\left(\frac{x}{1+x}\right)\right] = \frac{d}{dx}\left[\frac{x}{(1+x)^2}\right]\\ &= \frac{1-x}{(1+x)^3} \end{align} $$

achille hui
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$$ \sum_{n = 0}^\infty (-1)^n(n+1)^2x^n $$ should be exactly what you need, if I understand your question.

Arthur
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HINT: $$1-4x+9x^2-16x^3 + \cdots = \sum_{n=0}^{\infty} (-1)^n (n+1)^2 x^n = \sum_{n=0}^{\infty} (-1)^n (n^2+2n+1) x^n$$

SchrodingersCat
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    Why the downvote? – SchrodingersCat Nov 26 '15 at 13:11
  • @LTS: rudeness/arrogance is not welcome here. Many people could mock you for not being able to solve the easy question, if they wanted. –  Nov 26 '15 at 13:21
  • @YvesDaoust The people posting trivial answers couldn't, else they would actually solve it. – user85798 Nov 26 '15 at 13:26
  • @LTS Perhaps you are unable to understand the meaning of the word "HINT". – SchrodingersCat Nov 26 '15 at 13:29
  • @lts: I preferred to withdraw my answer. –  Nov 26 '15 at 13:30
  • @LTS And as far as solving the series is considered, if that was your question, then nobody would have even cared to do so for it is already a near duplicate. And please.. learn some decency and modesty... this is a public site, if you cannot respect others, I don't think anyone will answer your questions. – SchrodingersCat Nov 26 '15 at 13:32
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Multiply your expression by $(1+x)^3$ and find out.

mercio
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