## Precalculus (6th Edition) Blitzer

$\dfrac{1995}{64}$
As we are given that $\sum_{i=1}^{6}(\dfrac{3}{2})^i$ Now, terms will be {$\dfrac{3}{2},\dfrac{9}{4},\dfrac{27}{8}...$} This shows an geometric sequence. whose sum will be $S_n=\dfrac{a_1(1-r^n)}{1-r}$ Thus, $S_6=\dfrac{(\dfrac{3}{2})(1-(\dfrac{3}{2})^6)}{1-\dfrac{3}{2}}=\dfrac{\dfrac{3}{2}(\dfrac{-665}{64})}{-1/2}$ $S_6=\dfrac{1995}{64}$