To find the sum of the series given by [tex]\(\sum_{n=1}^5 4(3)^{n-1}\)[/tex], we can follow these steps:
1. Identify the terms of the series: The series follows the form [tex]\(4(3)^{n-1}\)[/tex] for each [tex]\(n\)[/tex] from 1 to 5. Let’s list out the terms:
- For [tex]\(n = 1\)[/tex]: [tex]\(4 \cdot 3^{1-1} = 4 \cdot 3^0 = 4 \cdot 1 = 4\)[/tex]
- For [tex]\(n = 2\)[/tex]: [tex]\(4 \cdot 3^{2-1} = 4 \cdot 3^1 = 4 \cdot 3 = 12\)[/tex]
- For [tex]\(n = 3\)[/tex]: [tex]\(4 \cdot 3^{3-1} = 4 \cdot 3^2 = 4 \cdot 9 = 36\)[/tex]
- For [tex]\(n = 4\)[/tex]: [tex]\(4 \cdot 3^{4-1} = 4 \cdot 3^3 = 4 \cdot 27 = 108\)[/tex]
- For [tex]\(n = 5\)[/tex]: [tex]\(4 \cdot 3^{5-1} = 4 \cdot 3^4 = 4 \cdot 81 = 324\)[/tex]
2. Sum the terms: Add up all the terms calculated above:
- [tex]\(4 + 12 + 36 + 108 + 324\)[/tex]
3. Calculate the total:
- [tex]\(4 + 12 = 16\)[/tex]
- [tex]\(16 + 36 = 52\)[/tex]
- [tex]\(52 + 108 = 160\)[/tex]
- [tex]\(160 + 324 = 484\)[/tex]
Hence, the sum of the series [tex]\(\sum_{n=1}^5 4(3)^{n-1}\)[/tex] is:
[tex]\[ \boxed{484} \][/tex]
