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

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TopNewest\(\displaystyle \frac{x_1^{2}}{1-x_1} + \frac{x_2^{2}}{1-x_2}+\ldots + \frac{x_{2014}^{2}}{1-x_{2014}}\)

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 · 2 years, 7 months ago

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– Aditya Kumar · 1 year ago

Exactly did the sameLog in to reply

– Pranshu Gaba · 2 years, 7 months ago

Nice !!!!!Log in to reply

just subtract 1st eqn from 2..u get the required 3rd eqn =1-1=0 – Incredible Mind · 2 years, 5 months ago

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0 – Anshuman Bais · 1 year, 10 months ago

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