Азия-тынық мұхит математикалық олимпиадасы, 2000 жыл

Вычислите сумму \[S=\sum\limits_{i=0}^{100}{\frac{x_{i}^{3}}{1-3{{x}_{i}}+3x_{i}^{2}}}\] для всех ${{x}_{i}}=\frac{i}{101}$.

Комментарий/решение

$$S=\sum_{i=0}^{100}\frac{x_i^3}{1-3x_i+3x_i^3}=\sum_{i=0}^{100}\frac{x_i^3}{(1-x_i)^3+x_i^3}=\sum_{i=0}^{100}\frac{i^3}{(101-i)^3+i^3}$$

$$a_i=\frac{i^3}{(101-i)^3+i^3}\Rightarrow a_i+a_{101-i}=1, \quad a_0=0$$

$$S=\sum_{i=0}^{100} a_i=\sum_{i=0}^{50}(a_i+a_{101-i})=\underbrace{1+1+...+1}_{50}=50$$

$1-x_i=1-{i\over 101}={{101-i}\over 101}=x_{101-i}$

\begin{eqnarray*}\sum_{i=0}^{101}\frac{x_i^3}{1-3x_i+3x_i^2} &=& \sum_{i=0}^{101}\frac{x_i^3}{(1-x_i)^3+x_i^3}\\ &=& \sum_{i=0}^{101}\frac{x_i^3}{x_{101-i}^3+x_i^3}\\ &=& \sum_{i=0}^{50}\left(\frac{x_i^3}{x_{101-i}^3+x_i^3}+\frac{x_{101-i}^3}{x_{101-i}^3+x_i^3}\right)\\ &=& \sum_{i=0}^{50} 1\\ &=& 51\end{eqnarray*}