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- Feb 14, 2012

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Find \(\displaystyle \sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}\).

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- Thread starter
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- #1

- Feb 14, 2012

- 3,909

Find \(\displaystyle \sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}\).

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- Feb 14, 2012

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Hey

You were so naughty and didn't want to play with this problem when I first asked you to solve it months ago!

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- Jan 25, 2013

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\(\displaystyle f(x)=(\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1})\times \dfrac {10201}{10201}Find \(\displaystyle \sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}\).

=\dfrac {202x-10201}{3x^2-303x+10201}\)

let y=101-x, then x=101-y

\(\displaystyle f(x)=\dfrac {202(101-y)-10201}{3(101-y)^2-303(101-y)+10201}=\dfrac {10201-202y}{3y^2-303y+10201}\)

$\therefore f(0)=-f(101), f(1)=-f(100),-------,f(50)=-f(51)$

that is :

f(0)+f(101)=f(1)+f(100)=f(2)+f(99)=----------=f(50)+f(51)=0

and we get :

\(\displaystyle \sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}=0\)