Problem
ALG-B1-M10-P010 Explicit formula
#10
★★☆☆☆ Level 2 of 5
Let \(u_0=0\), \(u_{n+1}=u_n+3n+1\). Find \(u_n\).
Sum \(3k+1\) from \(0\) to \(n-1\).
\(u_n=\sum_{k=0}^{n-1}(3k+1)=3\frac{n(n-1)}{2}+n=\frac{3n^2-n}{2}\).
Finite differences give a quadratic formula.