## College Algebra (10th Edition)

$$\sum_{k=1}^{n}(2k+1) = n^2+2n$$
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, $$\sum_{k=1}^{n}(2k+1) = n^2+2n$$