Answer
-18
Work Step by Step
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:
$\sum_{i=1}^6(2+i-i^2)
\\=\sum_{i=1}^62+\sum_{i=1}^6i-\sum_{i=1}^6i^2
\\=(6\times 2)+\frac{6\times 7}{2}-\frac{6\times 7\times 13}{6}
\\=12+21-91
\\=-58$
