Consider the series below:
[tex]\[ \sum_{k=1}^{11} 3\left(\frac{1}{6}\right)^{(k-1)} \][/tex]

Which of the following represent the first four terms in the given series?

A. [tex]\(3, \frac{3}{2}, \frac{9}{2}, \frac{27}{2}\)[/tex]

B. [tex]\(\frac{1}{2}, \frac{3}{2}, \frac{9}{2}, \frac{27}{2}\)[/tex]

C. [tex]\(3, \frac{1}{2}, \frac{1}{12}, \frac{1}{72}\)[/tex]

D. [tex]\(\frac{1}{2}, \frac{1}{12}, \frac{1}{72}, \frac{1}{432}\)[/tex]



Answer :

Let's analyze the series [tex]\( \sum_{k=1}^{11} 3\left(\frac{1}{6}\right)^{(k-1)} \)[/tex] by finding the first four terms in the series.

### Step-by-Step Calculation:

1. First term (k=1):
[tex]\[ \text{Term}_1 = 3\left(\frac{1}{6}\right)^{1-1} = 3\left(\frac{1}{6}\right)^0 = 3 \times 1 = 3 \][/tex]

2. Second term (k=2):
[tex]\[ \text{Term}_2 = 3\left(\frac{1}{6}\right)^{2-1} = 3\left(\frac{1}{6}\right)^1 = 3 \times \frac{1}{6} = \frac{3}{6} = 0.5 \][/tex]

3. Third term (k=3):
[tex]\[ \text{Term}_3 = 3\left(\frac{1}{6}\right)^{3-1} = 3\left(\frac{1}{6}\right)^2 = 3 \times \left(\frac{1}{6}\right)^2 = 3 \times \frac{1}{36} \approx 0.0833333333333333 \][/tex]

4. Fourth term (k=4):
[tex]\[ \text{Term}_4 = 3\left(\frac{1}{6}\right)^{4-1} = 3\left(\frac{1}{6}\right)^3 = 3 \times \left(\frac{1}{6}\right)^3 = 3 \times \frac{1}{216} \approx 0.01388888888888889 \][/tex]

From these calculations, we get the first four terms of the series as:
[tex]\[ (3, 0.5, 0.0833333333333333, 0.0138888888888889) \][/tex]

### Comparing with the Given Choices:

- Option A: [tex]\(3, \frac{3}{2}, \frac{9}{2}, \frac{27}{2}\)[/tex]
- Option B: [tex]\(\frac{1}{2}, \frac{3}{2}, \frac{9}{2}, \frac{27}{2}\)[/tex]
- Option C: [tex]\(3, \frac{1}{2}, \frac{1}{12}, \frac{1}{72}\)[/tex]
- Option D: [tex]\(\frac{1}{2}, \frac{1}{12}, \frac{1}{72}, \frac{1}{432}\)[/tex]

The first four terms we calculated are [tex]\( (3, 0.5, 0.0833333333333333, 0.0138888888888889) \)[/tex].

Clearly, the correct answer which matches our calculated terms is:
[tex]\[ \boxed{\text{C}} \][/tex]