MHB Sum of a Complex Fraction Sequence

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The sum $$\sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}$$ evaluates to zero due to the symmetry of the numerator and denominator about a specific value of x. The numerator is an odd function while the denominator is even, which aligns with the odd function rule in integral calculus. By substituting x with 101 - y, it is demonstrated that pairs of terms cancel each other out, leading to a total sum of zero. This symmetry is crucial in simplifying the evaluation of the sum. The conclusion is that the entire sum equals zero.
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Find $$\sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}$$.
 
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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.
 
MarkFL said:
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)
 
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)
 
anemone said:
Find $$\sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}$$.
$$f(x)=(\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1})\times \dfrac {10201}{10201}
=\dfrac {202x-10201}{3x^2-303x+10201}$$
let y=101-x, then x=101-y
$$f(x)=\dfrac {202(101-y)-10201}{3(101-y)^2-303(101-y)+10201}=\dfrac {10201-202y}{3y^2-303y+10201}$$
$\therefore f(0)=-f(101), f(1)=-f(100),-------,f(50)=-f(51)$
that is :
f(0)+f(101)=f(1)+f(100)=f(2)+f(99)=----------=f(50)+f(51)=0
and we get :
$$\sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}=0$$
 
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