shmilybaby4i

2022-06-27

Find the value of $\frac{{x}_{1}^{2}}{1-{x}_{1}}+\frac{{x}_{2}^{2}}{1-{x}_{2}}+...+\frac{{x}_{n}^{2}}{1-{x}_{n}}.$.
Suppose, and $\frac{{x}_{1}}{1-{x}_{1}}+\frac{{x}_{2}}{1-{x}_{2}}+...+\frac{{x}_{n}}{1-{x}_{n}}=1.$ Find the value of $\frac{{x}_{1}^{2}}{1-{x}_{1}}+\frac{{x}_{2}^{2}}{1-{x}_{2}}+...+\frac{{x}_{n}^{2}}{1-{x}_{n}}.$

candelo6a

Expert

$\frac{{x}_{1}^{2}}{1-{x}_{1}}+\frac{{x}_{2}^{2}}{1-{x}_{2}}+...+\frac{{x}_{n}^{2}}{1-{x}_{n}}=\frac{{x}_{1}^{2}}{1-{x}_{1}}+\frac{{x}_{2}^{2}}{1-{x}_{2}}+...+\frac{{x}_{n}^{2}}{1-{x}_{n}}-1+1\phantom{\rule{0ex}{0ex}}=\frac{{x}_{1}^{2}-{x}_{1}}{1-{x}_{1}}+\frac{{x}_{2}^{2}-{x}_{2}}{1-{x}_{2}}+...+\frac{{x}_{n}^{2}-{x}_{n}}{1-{x}_{n}}+1=-\left({x}_{1}+{x}_{2}+..+{x}_{n}\right)+1=0$

telegrafyx

Expert

Let $S$ be the sum of the series $\frac{{x}_{1}^{2}}{1-{x}_{1}}+\frac{{x}_{2}^{2}}{1-{x}_{2}}+\cdots +\frac{{x}_{n}^{2}}{1-{x}_{n}}$ , represented by $S=\sum _{i=1}^{n}\frac{{x}_{i}^{2}}{1-{x}_{i}}$
First, let's do some algebra.
$\begin{array}{rl}\frac{{x}_{i}^{2}}{1-{x}_{i}}& =\frac{{x}_{i}^{2}-2{x}_{i}+2{x}_{i}-1+1}{1-{x}_{i}}\\ & =\frac{\left({x}_{i}^{2}-2{x}_{i}+1\right)+\left({x}_{i}-1\right)+{x}_{i}}{1-{x}_{i}}\\ & =\frac{\left({x}_{i}-1{\right)}^{2}}{1-{x}_{i}}+\frac{{x}_{i}-1}{1-{x}_{i}}+\frac{{x}_{i}}{1-{x}_{i}}\\ & =\left(1-{x}_{i}\right)-1+\frac{{x}_{i}}{1-{x}_{i}}\\ & =\frac{{x}_{i}}{1-{x}_{i}}-{x}_{i}\end{array}$
Now we can simplify our sum.
$S=\sum _{i=1}^{n}\frac{{x}_{i}^{2}}{1-{x}_{i}}=\sum _{i=1}^{n}\left(\frac{{x}_{i}}{1-{x}_{i}}-{x}_{i}\right)=\sum _{i=1}^{n}\frac{{x}_{i}}{1-{x}_{i}}-\sum _{i=1}^{n}{x}_{i}=1-1=0$

Do you have a similar question?