Which shows the terms of the series? [tex]$\sum_{n=1}^4 3(4)^{n-1}$[/tex]
A. [tex]$4+12+48+192$[/tex]
B. [tex]$4+12+36$[/tex]
C. [tex]$3+12+48+192$[/tex]
D. [tex]$3+12+48$[/tex]
Answer (1)
To solve this problem, we need to understand what the given series ∑ n = 1 4 3 ( 4 ) n − 1 represents. This is a geometric series where the general term is defined as 3 ( 4 ) n − 1 . We need to find the specific terms from the first term to the fourth term.
Let's break it down step by step:
Calculate the first term ( n = 1 ): 3 ( 4 ) 1 − 1 = 3 ( 4 ) 0 = 3 ( 1 ) = 3
Calculate the second term ( n = 2 ): 3 ( 4 ) 2 − 1 = 3 ( 4 ) 1 = 3 ( 4 ) = 12
Calculate the third term ( n = 3 ): 3 ( 4 ) 3 − 1 = 3 ( 4 ) 2 = 3 ( 16 ) = 48
Calculate the fourth term ( n = 4 ): 3 ( 4 ) 4 − 1 = 3 ( 4 ) 3 = 3 ( 64 ) = 192
Now, list the terms calculated:
3 + 12 + 48 + 192
Looking at the given options, the correct answer is C. 3 + 12 + 48 + 192 .
