3

The answer is: $\frac{-2z^2}{(1-z)^{3}}$

When I do it, I get the answer without the minus sign.

So far, I got:

$$ \begin{split} \sum_{n=2}^{\infty}(n^2-3n+2)z^n &= \sum_{n=2}^{\infty}(n-1)(n-2)z^n \\ &= z^3 \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-1} \\ &= z^3 \frac{d^2}{dz^2} \sum_{n=1}^{\infty}z^{n-1} \end{split} $$

Andreas
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Opay Tress
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  • I've seen this exact question asked at least three times in the last 24 hours. – MSDG Nov 01 '18 at 13:04
  • I think you should roll back that edit, but the previous version leads me to conjecture the sign error it mentioned resulted from you making a mistake in either the first or second differentiation, starting at $1/(1-z)$. Either that, or you mistakenly started the sum in such a place as to think you were differentiating $z/(1-z)$ twice. – J.G. Nov 01 '18 at 21:15
  • @Opay Tress: please don't edit the question in a way that Significantly change the post, deleting the effort you showed originally is also no good. – ℋolo Nov 01 '18 at 21:16

1 Answers1

2

\begin{split} \sum_{n=2}^{\infty}(n^2-3n+2)z^{n-1} &= \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-1} \\ &= z^2 \sum_{n=2}^{\infty}(n-1)(n-2)z^{n-3} \\ &= z^2 \frac{d^2}{dz^2} \sum_{n=1}^{\infty}z^{n-1}\\ &= z^2 \frac{d^2}{dz^2} \sum_{n=0}^{\infty}z^{n}\\ &= z^2 \frac{d^2}{dz^2} \frac{1}{1-z}\\ &= z^2 \frac{2}{(1-z)^3} \end{split}

This proof holds for all complex $z$ with $|z|<1$ which means that, in particular, this also holds for all real $z$ with $0<z<1$. From the task setting itself, in this case the answer must be positive, since only positive terms are added. So if the result is claimed to be $ \frac{-2 z^2}{(1-z)^3}$, this is false.

Andreas
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  • A question that has nothing to do with this one : we had answered, both, about 3-4 days ago to a question dealing with a system of equations with 6 variables $x_1,x_2,x_3,y_1,y_2,y_3$ ; this question and its answers are lost from the radar screen. may I ask you if you know why ? – Jean Marie Nov 02 '18 at 16:15
  • @JeanMarie The last modification I had seen about this question was that moderators soft-deleted it because of MSE rule-violation: the question is contained in an ongoing math contest. I guess that the author now has fully removed the question. – Andreas Nov 02 '18 at 16:57
  • I see. Thank you very much for taking time to answer me. – Jean Marie Nov 02 '18 at 17:11