Answer:
Option 3 rd is correct
sum of the given geometric series is, 40
Step-by-step explanation:
The sum of the finite geometric series is given by:
[tex]S_n=\frac{a_1(1-r^n)}{1-r}[/tex] [tex]r\neq 1[/tex] ....[1]
Given that:
[tex]\sum_{n=1}^{4} (-2)(-3)^{n-1}[/tex]
We have to find the [tex]S_{4}[/tex] for the series:
[tex](-2)(-3)^0+(-2)(-3)^1+(-2)(-3)^2+(-2)(-3)^3[/tex]
[tex]-2+6-18+54[/tex]
here, [tex]a_1 = -2[/tex] and common ratio(r) = -3
Substitute these in [1] we have;
[tex]S_4=\frac{-2 \cdot (1-(-3)^{4})}{1-(-3)}[/tex]
⇒ [tex]S_4=\frac{-2 \cdot (1-(81))}{1+3}[/tex]
⇒ [tex]S_4=\frac{-2 \cdot -80}{4}[/tex]
Simplify:
[tex]S_4 =\frac{160}{4} = 40[/tex]
therefore, the sum of the given geometric series is, 40
