Answer
$\sum\limits_{i =1}^{n}(i^{3}-i-2)=\frac{1}{4}n(n^{3}+2n^{2}-n-10)$
Work Step by Step
Find the value of the sum $\sum\limits_{i =1}^{n}(i^{3}-i-2)$
After expanding the terms, we have
$\sum\limits_{i =1}^{n}(i^{3}-i-2)=\sum\limits_{i =1}^{n}i^{3}-\sum\limits_{i =1}^{n}i-2\sum\limits_{i =1}^{n}1$
$=[\frac{n(n+1)}{2}]^{2}-[\frac{n(n+1)}{2}]-2n$
$=\frac{1}{4}n^{2}(n+1)^{2}+\frac{1}{2}n(n+1)-2n$
$=\frac{1}{4}n[n(n+1)^{2}-2(n+1)-8]$
Hence, $\sum\limits_{i =1}^{n}(i^{3}-i-2)=\frac{1}{4}n(n^{3}+2n^{2}-n-10)$
