Answer
(a) $\frac{(1+n)n}{2}$
(b) $n+(1+n)3 n$
(c) $\frac{n(1+n)(1+2 n)}{6}$
Work Step by Step
Using theorem 5.4 .2:
(a) $\sum_{k=1}^{n} k=\frac{(1+n)n}{2}$
Using theorems 5.4 .1 and 5.4 .2:
(b) $\sum_{k=1}^{n}(1+6 k)=6 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$
$=6 \frac{(1+n)n}{2}+n=n+(1+n)3 n$
$(\mathrm{c}) \sum_{k=1}^{n} k^{2}=\frac{(1+n)(1+2 n)n}{6}$