Sum of a Complex Fraction Sequence

[anemone]

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

 

[MarkFL]

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.

 

[anemone]

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]

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)

 

[Albert1]

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