# Evaluate a sum

#### anemone

##### MHB POTW Director
Staff member
Find $$\displaystyle \sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}$$.

#### MarkFL

Staff member
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

##### MHB POTW Director
Staff member
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!

#### MarkFL

Staff member
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.

#### Albert

##### Well-known member
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 :
$$\displaystyle \sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}=0$$