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Evaluate a sum

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anemone

MHB POTW Director
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Feb 14, 2012
3,756
Find \(\displaystyle \sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}\).
 

MarkFL

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Feb 24, 2012
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The given sum is zero because the numerator and denominator are symmetric about the same value of x, with the numerator being odd and the denominator even with respect to this axis of symmetry. This is analogous to the odd function rule from integral calculus.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
The given sum is zero because the numerator and denominator are symmetric about the same value of x, with the numerator being odd and the denominator even with respect to this axis of symmetry. This is analogous to the odd function rule from integral calculus.
Hey MarkFL,

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I was probably having a "bad math day" then, as it was pretty straightforward tonight to simply look at the symmetry of the summand with respect to the index of summation. While I don't recall you asking me about this before, perhaps seeing it in $\LaTeX$ made a difference too. (Mmm)
 

Albert

Well-known member
Jan 25, 2013
1,225
Find \(\displaystyle \sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}\).
\(\displaystyle f(x)=(\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1})\times \dfrac {10201}{10201}
=\dfrac {202x-10201}{3x^2-303x+10201}\)
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 :
\(\displaystyle \sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}=0\)