what is the sum of the geometric series?\n sum_{n = 1}^{4}(-2)(-3)^{n - 1} \n-122\n-2\n40\n54

Answer:

C. 40

Explanation:

Step1: Identify the formula for sum of geometric series

The sum of a finite geometric series is given by $S_n=\sum_{k = 1}^{n}a\cdot r^{k - 1}=a\frac{1 - r^{n}}{1 - r}$, where $a$ is the first - term and $r$ is the common ratio, and $n$ is the number of terms. For the series $\sum_{n = 1}^{4}(-2)(-3)^{n - 1}$, we have $a=-2$, $r=-3$, and $n = 4$.

Step2: Substitute the values into the formula

[ \begin{align*} S_4&=(-2)\frac{1-(-3)^{4}}{1-(-3)}\ &=(-2)\frac{1 - 81}{1 + 3}\ &=(-2)\frac{-80}{4}\ &=(-2)\times(-20)\ &=40 \end{align*} ]