Answer:
Option (3)
Step-by-step explanation:
Given expression in this question represents the partial sum of an infinite geometric series in the sigma notation.
[tex]S_{n}=\sum_{n=1}^{\infty}6(2)^{n-1}[/tex]
First term of this series 'a' = 6
Common ratio 'r' = 2
We have to find the sum of 4 terms of this infinite series (n = 4).
Sum of n terms of a geometric series is,
[tex]S_n=\frac{a(r^n-1)}{(r-1)}[/tex]
[tex]S_4=\frac{6(2^4-1)}{(2-1)}[/tex]
[tex]=\frac{6(16-1)}{(1)}[/tex]
[tex]=90[/tex]
Therefore, sum of 4 terms of the given series will be 90.
Option (3) will be the answer.
