Thursday, December 1st

Problem
Find
$S=\sum_{i=0}^{101} \frac{x_i^3}{1-3x_i+3x_i^2}$
where $x_i=i/101$.


Solution
Since
$\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{(1-x_i)^3+x_i^3}$

is symmetric when exchanging $x_i$ by $1-x_i$, we have that every such pair will contribute to the sum by one, and there are 51 such pairs, hence $S=51$.