Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises - Page 615: 63

$S_5=\frac{211}{27}$

We need to find: $\sum_{k=1}^{5}3(\frac{2}{3})^{k-1}$ We see that this is a geometric sequence with $a=3$ and $r=\frac{2}{3}$. We know the partial sum of a geometric sequence is: $S_n=a_1\frac{1-r^n}{1-r}$ $S_5=3\frac{1-(\frac{2}{3})^{5}}{1-\frac{2}{3}}=\frac{211}{27}$