Answer
$$\sum_{k=1}^{n}(k+2) = \dfrac{n(n+1)}{2} + 2n$$
Work Step by Step
RECALL:
For any constant $c$,
(1) $$\sum_{k=1}^{n}(k+c) = \sum_{k=1}^{n}k + \sum_{k=1}^{n}c$$
(2) $$\sum_{i=1}^{n}c = nc$$
(3) $$\sum_{k=1}^{n}k = \dfrac{n(n+1)}{2}$$
Using rule (1) above gives:
$$\sum_{k=1}^n(k + 2) = \sum_{k=1}^{n}k + \sum_{k=1}^{n}2$$
Use rule (3) and rule (2) above, respectively, to obtain:
$$\sum_{k=1}^{n}k + \sum_{k=1}^{n}2 = \dfrac{n(n+1)}{2} + 2n$$
Thus,
$$\sum_{k=1}^{n}(k+2) = \dfrac{n(n+1)}{2} + 2n$$
