## Calculus: Early Transcendentals 8th Edition

$\sum\limits_{i =1}^{n}(i+1)(i+2)=\frac{1}{3}n(n^{2}+6n+11)$
Find the value of the sum $\sum\limits_{i =1}^{n}(i+1)(i+2)$ After expanding the terms, we have $\sum\limits_{i =1}^{n}(i+1)(i+2)=\sum\limits_{i =1}^{n}(i^{2}+3i+2)$ $=\sum\limits_{i =1}^{n}i^{2}+3\sum\limits_{i =1}^{n}i+\sum\limits_{i =1}^{n}2$ $=[\frac{n(n+1)(2n+1)}{6}]+3[\frac{n(n+1)}{2}]+2n$ $=\frac{1}{3}n^{3}+2n^{2}+\frac{11}{3}n$ Hence, $\sum\limits_{i =1}^{n}(i+1)(i+2)=\frac{1}{3}n(n^{2}+6n+11)$