Answer
$$\sum_{k=1}^{n}(2k+1) = n^2+2n$$
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) $$\sum_{k=1}^{n}k = \dfrac{n(n+1)}{2}$$
(4) $$\sum_{k=1}^{n}ck= c\sum_{k=1}^{n}k$$
Using rule (1) above gives:
$$\sum_{k=1}^n(2k + 1) = \sum_{k=1}^{n}2k + \sum_{k=1}^{n}1$$
Using rule (4) above gives:
$$2\sum_{k=1}^{n}k + \sum_{k=1}^{n}1$$
Use rule (3) and rule (2) above, respectively, to obtain:
$$2\sum_{k=1}^{n}k + \sum_{k=1}^{n}1
\\= 2\left(\dfrac{n(n+1)}{2}\right) + 1(n)
\\= n(n+1) + n
\\=n^2+n+n
\\= n^2+2n$$
Thus,
$$\sum_{k=1}^{n}(2k+1) = n^2+2n$$
