Pre-RMO 2014/19

Let $x_1, x_2, \ldots, x_{2014}$ be real numbers different from $1$ such that $x_1 + x_2 + \ldots + x_{2014} = 1$ and

$\frac{x_1}{1 - x_1} + \frac{x_2}{1 - x_2} + \ldots + \frac{x_{2014}}{1 - x_{2014}} = 1.$

What is the value of

$\frac{x^2_1}{1 - x_1} + \frac{x^2_2}{1 - x_2} + \frac{x^2_3}{1 - x_3} + \ldots + \frac{x^2_{2014}}{1 - x_{2014}} ?$

This note is part of the set Pre-RMO 2014

#Algebra #Pre-RMO

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Comments

$\displaystyle \frac{x_1^{2}}{1-x_1} + \frac{x_2^{2}}{1-x_2}+\ldots + \frac{x_{2014}^{2}}{1-x_{2014}}$

Now let us add

$\displaystyle x_1 + x_2 + \ldots + x_{2014}$

To the above equation

$\displaystyle \frac{x_1^{2}}{1-x_1} + x_1 + \frac{x_2^{2}}{1-x_2} + x_2 + \ldots \frac{x_{2014}^{2}}{1-x_{2014}} + x_{2014}$

Now taking L.C.M

We finally get

$\displaystyle \frac{x_1}{1-x_1} +\frac{x_2}{1-x_2} + \ldots \frac{x_{2014}}{1-x_{2014}}$

So this is equal to 1 but we have added

$x_1 + x_2 \ldots + x_{2014 }= 1$

So we have to subtract 1

Finally we get the answer $\boxed {0}$

Krishna Sharma - 6 years, 8 months ago

Nice !!!!!

Pranshu Gaba - 6 years, 8 months ago

Exactly did the same

Aditya Kumar - 5 years ago

just subtract 1st eqn from 2..u get the required 3rd eqn =1-1=0

incredible mind - 6 years, 6 months ago

Anshuman Singh Bais - 5 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$