## College Algebra (11th Edition)

The summation can be split up based on its properties. If k is a constant, $\sum_{i=1}^n(k\times (a_{i}))=k\times \sum_{i=1}^n(a_{i})$ If k is a constant, $\sum_{i=1}^nk=n\times k$ $\sum_{i=1}^n(a_{i}+b_{i})=\sum_{i=1}^n(a_{i})+\sum_{i=1}^n(b_{i})$ All properties above require n to be a positive integer and $a_{i}$ and $b_{i}$ are two sequences. Here, $n=6$ $\sum_{i=1}^6(2+i-i^2)=\sum_{i=1}^62+\sum_{i=1}^6i-\sum_{i=1}^6i^2$ The summation of square numbers: $\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}$ The summation of positive integers: $\sum_{i=1}^ni=\frac{n(n+1)}{2}$ By substituting these summations into the given sum, with $n=6$ we get: $\sum_{i=1}^6(2+i-i^2) \\=\sum_{i=1}^62+\sum_{i=1}^6i-\sum_{i=1}^6i^2 \\=(6\times 2)+\frac{6\times 7}{2}-\frac{6\times 7\times 13}{6} \\=12+21-91 \\=-58$