Question

Write the series represented by
n=1
∑
6
​
(8)(
2
1
​
)
n−1

Answer 1

To evaluate the series represented by

\[
\sum_{n=1}^{6} 8 \left( \frac{2}{1} \right)^{n-1},
\]

let's break it down step by step.

1. **Identify the terms of the series:**
The general term of the series can be expressed as:
\[
8 \left( \frac{2}{1} \right)^{n-1} = 8 \cdot 2^{n-1}.
\]

2. **Write out the terms of the series from \( n = 1 \) to \( n = 6 \):**
- For \( n = 1 \): \( 8 \cdot 2^{1-1} = 8 \cdot 2^0 = 8 \cdot 1 = 8 \)
- For \( n = 2 \): \( 8 \cdot 2^{2-1} = 8 \cdot 2^1 = 8 \cdot 2 = 16 \)
- For \( n = 3 \): \( 8 \cdot 2^{3-1} = 8 \cdot 2^2 = 8 \cdot 4 = 32 \)
- For \( n = 4 \): \( 8 \cdot 2^{4-1} = 8 \cdot 2^3 = 8 \cdot 8 = 64 \)
- For \( n = 5 \): \( 8 \cdot 2^{5-1} = 8 \cdot 2^4 = 8 \cdot 16 = 128 \)
- For \( n = 6 \): \( 8 \cdot 2^{6-1} = 8 \cdot 2^5 = 8 \cdot 32 = 256 \)

3. **Sum the terms:**
Adding these values together:
\[
8 + 16 + 32 + 64 + 128 + 256.
\]

Calculating step by step:
- \( 8 + 16 = 24 \)
- \( 24 + 32 = 56 \)
- \( 56 + 64 = 120 \)
- \( 120 + 128 = 248 \)
- \( 248 + 256 = 504 \)

4. **Final result:**
Therefore, the sum of the series is:

\[
\sum_{n=1}^{6} 8 \left( \frac{2}{1} \right)^{n-1} = 504.
\]