Question

What is the sum of the geometric series: 

Answer 1

by looking at the expression, we can pretty much see who's the common factor and first term.

$\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \sum\limits_{n=1}^{4}~(\stackrel{\stackrel{a_1}{\downarrow }}{-2})(\stackrel{\stackrel{r}{\downarrow }}{-3})^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\[-0.5em] \hrulefill\\ a_1=-2\\ r=-3\\ n=4 \end{cases} \\\\\\ S_4=-2\left( \cfrac{1-(-3)^4}{1-(-3)} \right)\implies S_4=-2\left( \cfrac{1-81}{1+3} \right) \\\\\\ S_4=-2\cdot \cfrac{-80}{4}\implies S_4=-\cfrac{-80}{2}\implies S_4=40$

Answer 2

Answer:

Option C on edge :)

Step-by-step explanation: