Write the series represented by ∑6n=1(8)(12)n−1

Answer 1

The series you provided is \(\sum_{n=1}^{6} (8)(12)^{n-1}\).

This is a finite series where the first term can be calculated when \(n=1\) and the last term when \(n=6\). Let's calculate the first few terms to understand the series better:

1. For \(n = 1\): \((8)(12)^{1-1} = 8 \cdot 1 = 8\)
2. For \(n = 2\): \((8)(12)^{2-1} = 8 \cdot 12 = 96\)
3. For \(n = 3\): \((8)(12)^{3-1} = 8 \cdot 12^2 = 8 \cdot 144 = 1152\)
4. For \(n = 4\): \((8)(12)^{4-1} = 8 \cdot 12^3 = 8 \cdot 1728 = 13824\)
5. For \(n = 5\): \((8)(12)^{5-1} = 8 \cdot 12^4 = 8 \cdot 20736 = 165888\)
6. For \(n = 6\): \((8)(12)^{6-1} = 8 \cdot 12^5 = 8 \cdot 248832 = 1990656\)

Putting it all together, the series can be expressed as:

\[ 8 + 96 + 1152 + 13824 + 165888 + 1990656 \]

So, the complete series is:

\[ \sum_{n=1}^{6} (8)(12)^{n-1} = 8 + 96 + 1152 + 13824 + 165888 + 1990656 \]