Question

Find S4 of a1=64, an=-3/4an-1, n>2

Answer 1

just to clarify some part of the notation

`a_n=-\cfrac{3}{4}a_(n-1)` this is simply saying that whatever "nth" term you wish to get, is -3/4 times whatever is before it, so another wording for that will be, the next term is the current term times -3/4, so in short -3/4 is the common ratio.

`\qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle S_n=\sum\limits_(i=1)^(n)\ a_1\cdot r^(i-1)\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\[-0.5em] \hrulefill\\ a_1=64\\ n=4\\ r=-(3)/(4) \end{cases}`

`S_4=64\left( \cfrac{1-\left( -(3)/(4) \right)^4}{1-\left( -(3)/(4) \right)} \right)\implies S_4=64\left( \cfrac{~~ ( 175 )/( 256 ) ~~}{(7)/(4)} \right) \\\\\\ S_4=64\left( \cfrac{25}{64} \right)\implies S_4=25`