Answer:
1456
Step-by-step explanation:
This question is asking us to find the sum of [tex]4(3)^{n-1}[/tex] from n = 1 to n = 6. We can find the sum like this:
[tex]4(3)^{1-1} +4(3)^{2-1} +4(3)^{3-1} +4(3)^{4-1} +4(3)^{5-1} +4(3)^{6-1}[/tex]
Which can be simplified to:
[tex]4(3)^{0} +4(3)^{1} +4(3)^{2} +4(3)^{3} +4(3)^{4} +4(3)^{5}[/tex]
[tex]4(1) +4(3) +4(9) +4(27)+4(81) +4(243)[/tex]
[tex]4 +12 +36 +108+324 +972 = 1456[/tex]
