a_n = \frac1n \cdot \fracn(n+1)2 - \frac13n^3 \cdot \left( \fracn(n+1)2 \right)^2 - AMAZONAWS

📅 March 11, 2026 👤 scraface

Mar 11, 2026

$a_n = \frac{1}{n} \cdot \frac{n(n+1)}{2} - \frac{1}{3n^3} \cdot \left( \frac{n(n+1)}{2} \right)^2$