The sum to infinity of the series `1+2/3+6/(3^2)+(10)/(3^3)+(14)/(3^4). . . . . .` is (1) 2 (2) 3 (3) 4 (4) 6

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To find the sum to infinity of the series \( S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \ldots \), we can analyze the series step by step.

### Step 1: Identify the series
The series can be expressed as:
\[ S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \ldots \]
We notice that the denominators form a geometric progression (GP) with a common ratio of \( \frac{1}{3} \).

### Step 2: Analyze the numerators
The numerators are \( 1, 2, 6, 10, 14 \). Let's find the differences:
- \( 2 - 1 = 1 \)
- \( 6 - 2 = 4 \)
- \( 10 - 6 = 4 \)
- \( 14 - 10 = 4 \)

The first difference is \( 1 \), and the subsequent differences are constant at \( 4 \). This suggests that the numerators form a quadratic sequence.

### Step 3: General term of the numerator
The \( n \)-th term of the numerator can be expressed as:
\[ a_n = n^2 + n \]
This can be verified:
- For \( n=1 \): \( 1^2 + 1 = 2 \)
- For \( n=2 \): \( 2^2 + 2 = 6 \)
- For \( n=3 \): \( 3^2 + 3 = 12 \)
- For \( n=4 \): \( 4^2 + 4 = 20 \)

### Step 4: Rewrite the series
Thus, we can rewrite the series as:
\[ S = \sum_{n=0}^{\infty} \frac{n^2 + n}{3^n} \]

### Step 5: Split the series
We can split this into two separate series:
\[ S = \sum_{n=0}^{\infty} \frac{n^2}{3^n} + \sum_{n=0}^{\infty} \frac{n}{3^n} \]

### Step 6: Use known formulas
We can use the formulas for the sums of these series:
1. The sum of the series \( \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \) for \( |x| < 1 \).
2. The sum \( \sum_{n=0}^{\infty} n x^n = \frac{x}{(1-x)^2} \).
3. The sum \( \sum_{n=0}^{\infty} n^2 x^n = \frac{x(1+x)}{(1-x)^3} \).

For \( x = \frac{1}{3} \):
- \( \sum_{n=0}^{\infty} n x^n = \frac{\frac{1}{3}}{(1 - \frac{1}{3})^2} = \frac{\frac{1}{3}}{(\frac{2}{3})^2} = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4} \)
- \( \sum_{n=0}^{\infty} n^2 x^n = \frac{\frac{1}{3}(1 + \frac{1}{3})}{(1 - \frac{1}{3})^3} = \frac{\frac{1}{3} \cdot \frac{4}{3}}{(\frac{2}{3})^3} = \frac{\frac{4}{9}}{\frac{8}{27}} = \frac{4 \cdot 27}{9 \cdot 8} = \frac{12}{8} = \frac{3}{2} \)

### Step 7: Combine the results
Now we can combine the results:
\[ S = \frac{3}{2} + \frac{3}{4} = \frac{6}{4} + \frac{3}{4} = \frac{9}{4} \]

### Step 8: Final calculation
The final sum \( S \) is:
\[ S = 3 \]

Thus, the sum to infinity of the series is \( \boxed{3} \).