Suppose, $x_1+x_2+...+x_n=1\space (x_i\in \mathbb{R}, x_i ≠1)$ and $\frac{ x_1}{1-x_1}+\frac{x_2}{1-x_2}+...+\frac{x_n}{1-x_n}=1.$$\\$ Find the value of $\frac{ x_1^2}{1-x_1}+\frac{x_2^2}{1-x_2}+...+\frac{x_n^2}{1-x_n}.$
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$\frac{ x_1^2}{1-x_1}+\frac{x_2^2}{1-x_2}+...+\frac{x_n^2}{1-x_n}=\frac{ x_1^2}{1-x_1}+\frac{x_2^2}{1-x_2}+...+\frac{x_n^2}{1-x_n} - 1 +1\\=\frac{ x_1^2 -x_1}{1-x_1}+\frac{x_2^2 - x_2}{1-x_2}+...+\frac{x_n^2-x_n}{1-x_n} + 1=-(x_1 + x_2 + .. +x_n) + 1 = 0$
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I don't understand the down-vote. All that is wrong with your answer is that you've forgotten $+1$ in the third formula. – Rob Arthan Aug 25 '15 at 13:39
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@Rob Arthan I have a "friend" here. Seconds after my answers, I found myself down-voted, every time – Aug 25 '15 at 13:41
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oh dear! At least your answer is now impeccable. – Rob Arthan Aug 25 '15 at 13:47
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@gebra why not explaining the down-vote? – Aug 25 '15 at 13:49
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@Rob Arthan Thank you for correction. – Aug 25 '15 at 13:49
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Excuse me, but I'm not downvoting, my reputation isn't even high enough. – jg mr chapb Aug 25 '15 at 13:55
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2@gebra Sorry, I got it wrong. Let's forget it. – Aug 25 '15 at 13:56
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@EugenCovaci Let the moderators of SE know about your issue. On the other hand, maybe it is not a "friend", but the community telling you to stop giving non-pedagogical answers to copy-pasted homework-like questions: http://meta.math.stackexchange.com/questions/17057/answering-homework-questions-and-downvoting – Rol Aug 25 '15 at 15:05
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@Rol I got your point, but I still believe any down-vote must be explained – Aug 25 '15 at 15:18
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@EugenCovaci: i very much agree with you that down-votes should be explained. Roi is talking about "the community", but there was only one "down-voter". If that down-voter was a responsible pedagogue, then it is inexcusable that he or she did not give an explanation for the down-vote. The impression given by Roi is that there is a bunch of teachers out there who are out to kill any outsider who accidentally takes an interest in an MSE question and in so doing unintentionally gives away a solution to a homework problem. – Rob Arthan Aug 25 '15 at 21:22
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Let $S$ be the sum of the series ${x_1^2 \over 1 - x_1} + {x_2^2 \over 1 - x_2} + \cdots + {x_n^2 \over 1 - x_n}$ , represented by $S = \sum_{i=1}^n {x_i^2 \over 1-x_i}$
First, let's do some algebra. $$\begin{align*} {x_i^2 \over 1-x_i} & = {x_i^2 - 2x_i + 2x_i - 1 + 1 \over 1-x_i} \\ & = {(x_i^2 - 2x_i + 1) + (x_i - 1) + x_i \over 1-x_i} \\ & = {(x_i - 1)^2 \over 1-x_i} + {x_i - 1 \over 1-x_i} + {x_i \over 1-x_i} \\ & = (1 - x_i) - 1 + {x_i \over 1 -x_i} \\ & = {x_i \over 1-x_i} - x_i \\ \end{align*}$$
Now we can simplify our sum.
$$S = \sum_{i=1}^n {x_i^2 \over 1-x_i} = \sum_{i=1}^n ({x_i \over 1-x_i} - x_i)=\sum_{i=1}^n {x_i \over 1-x_i} - \sum_{i=1}^n x_i = 1 - 1 = 0$$
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