Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$3+4+5+6+....+(n+2)$
We see that there are $n$ terms, as the index $k$ changes from $1$ to $n$. The index $k$ indicates how the terms are formed. We write out the sum for the $n$ terms as follows: $\displaystyle \sum_{k=1}^{n} (k+2) = (1+2)+(2+2)+(3+2)+...+(n+2) \\=3+4+5+6+...+(n+2)$