## Precalculus (10th Edition)

In Problem 98, we proved: $u_{n+1}=\dfrac{(n+1)(n+2)}{2}$ Compute $u_{n+1}+u_n$ using the formula: $u_{n+1}+u_n=\dfrac{(n+1)(n+2)}{2}+\dfrac{n(n+1)}{2}$ $=\dfrac{n+1}{2}(n+2+n)$ $=\dfrac{(n+1)(2n+2)}{2}$ $=\dfrac{2(n+1)(n+1)}{2}$ $=(n+1)^2$ We proved: $u_{n+1}+u_n=(n+1)^2$