Find the value of $\frac{ x_1^2}{1-x_1}+\frac{x_2^2}{1-x_2}+...+\frac{x_n^2}{1-x_n}.$

Question

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 $\frac{ x_1^2}{1-x_1}+\frac{x_2^2}{1-x_2}+...+\frac{x_n^2}{1-x_n}.$

Answer 1

HINT:

$$a-\dfrac a{1-a}=\dfrac{a(1-a)-a}{1-a}=?$$

Answer 2

$\frac{ x_1^2}{1-x_1}+\frac{x_2^2}{1-x_2}+...+\frac{x_n^2}{1-x_n}=\frac{ x_1^2}{1-x_1}+\frac{x_2^2}{1-x_2}+...+\frac{x_n^2}{1-x_n} - 1 +1\\=\frac{ x_1^2 -x_1}{1-x_1}+\frac{x_2^2 - x_2}{1-x_2}+...+\frac{x_n^2-x_n}{1-x_n} + 1=-(x_1 + x_2 + .. +x_n) + 1 = 0$

Answer 3

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.

$$S = \sum_{i=1}^n {x_i^2 \over 1-x_i} = \sum_{i=1}^n ({x_i \over 1-x_i} - x_i)=\sum_{i=1}^n {x_i \over 1-x_i} - \sum_{i=1}^n x_i = 1 - 1 = 0$$