Answer
$\sum\limits_{i =1}^{n}(i+1)(i+2)=\frac{1}{3}n(n^{2}+6n+11)$
Work Step by Step
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, $\sum\limits_{i =1}^{n}(i+1)(i+2)=\frac{1}{3}n(n^{2}+6n+11)$