Problem 1

Compute the sum

S = \sum_{i=0}^{101} \frac{x_i^3}{1 - 3x_i + 3x_i^2}

for

x_i = \frac{i}{101}

Step 3 of 4: Reverse the summation order and add the two sums

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In plain words

Just like Gauss's trick for summing an arithmetic progression, pairing the $i$ -th term from the start with the $i$ -th term from the end makes every pair collapse to $1$ .

2S = \sum_{i=0}^{101} \frac{x_i^3}{x_{101-i}^3 + x_i^3} + \sum_{i=0}^{101} \frac{x_{101-i}^3}{x_i^3 + x_{101-i}^3} = \sum_{i=0}^{101} \frac{x_i^3 + x_{101-i}^3}{x_{101-i}^3 + x_i^3}

Detailed analysis

Replacing the dummy index $i$ by $101 - i$ in the sum $S = \sum_{i=0}^{101} \frac{x_i^3}{x_{101-i}^3 + x_i^3}$ gives $S = \sum_{i=0}^{101} \frac{x_{101-i}^3}{x_i^3 + x_{101-i}^3}$ . Adding these two expressions term by term combines the numerators over the common denominator $x_{101-i}^3 + x_i^3$ .