## Precalculus: Mathematics for Calculus, 7th Edition

See p. 855. For the arithmetic sequence $a_{n}=a+(n-1)d$ the nth partial sum $S_{n}=\displaystyle \sum_{k=1}^{n}[a+(k-1)d]$ is given by either of the following equivalent formulas: 1. $S_{n}=\displaystyle \frac{n}{2}[2a+(n-1)d]\qquad$2. $S_{n}=n(\displaystyle \frac{a+a_{n}}{2})$ ------------------- $\displaystyle \sum_{k=0}^{10}(3+0.25k)$. is a partial sum of an arithmetic series for which the first term is $a= 3+0.25\cdot 0=3$ , and $d=0.25$. There are n=11 terms in the sum (for k=0,1,2,...,10) The last term is $a_{11}=3+0.25\cdot 10=5.5$. Using formula 2, $S_{11}=\displaystyle \frac{11}{2}(3+5.5)=46.75$.