For a sequence `{a_(n)}, a_(1) = 2` and `(a_(n+1))/(a_(n)) = 1/3`, Then `sum_(r=1)^(oo) a_(r)` is

Options

A. `2//3`
B. `3`
C. `3//2`
D. prime number

Answer

To solve the problem, we need to find the sum of the sequence defined by the recurrence relation and the initial term provided. Let's break it down step by step.

### Step 1: Define the sequence
We are given:
- \( a_1 = 2 \)
- \( \frac{a_{n+1}}{a_n} = \frac{1}{3} \)

From this, we can express \( a_{n+1} \) in terms of \( a_n \):
\[ a_{n+1} = \frac{1}{3} a_n \]

### Step 2: Find the terms of the sequence
Using the recurrence relation, we can find the first few terms of the sequence:

- For \( n = 1 \):
\[ a_2 = \frac{1}{3} a_1 = \frac{1}{3} \times 2 = \frac{2}{3} \]

- For \( n = 2 \):
\[ a_3 = \frac{1}{3} a_2 = \frac{1}{3} \times \frac{2}{3} = \frac{2}{3^2} = \frac{2}{9} \]

- For \( n = 3 \):
\[ a_4 = \frac{1}{3} a_3 = \frac{1}{3} \times \frac{2}{9} = \frac{2}{3^3} = \frac{2}{27} \]

Continuing this pattern, we can see that:
\[ a_n = \frac{2}{3^{n-1}} \]

### Step 3: Write the sum of the series
We need to find the sum:
\[ \sum_{r=1}^{\infty} a_r = \sum_{r=1}^{\infty} \frac{2}{3^{r-1}} \]

This can be rewritten as:
\[ \sum_{r=0}^{\infty} \frac{2}{3^r} \]

### Step 4: Identify the sum of the geometric series
The series \( \sum_{r=0}^{\infty} x^r \) converges to \( \frac{1}{1-x} \) for \( |x| < 1 \). Here, \( x = \frac{1}{3} \):
\[ \sum_{r=0}^{\infty} \left(\frac{1}{3}\right)^r = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \]

Thus, multiplying by 2 (since we have \( 2 \) in front):
\[ \sum_{r=0}^{\infty} \frac{2}{3^r} = 2 \times \frac{3}{2} = 3 \]

### Final Answer
The sum \( \sum_{r=1}^{\infty} a_r \) is:
\[ \boxed{3} \]