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

Options

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

Answer

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 follow these steps:

### Step 1: Identify the pattern in the series
The numerators of the series are \( 1, 2, 6, 10, 14, \ldots \). We can observe that these numbers can be expressed in terms of a quadratic sequence. The \( n \)-th term can be represented as \( a_n = \frac{n(n+1)}{2} \) (the \( n \)-th triangular number).

### Step 2: Rewrite the series
The series can be rewritten as:
\[ S = \sum_{n=0}^{\infty} \frac{a_n}{3^n} = \sum_{n=0}^{\infty} \frac{n(n+1)/2}{3^n} \]
This can be simplified to:
\[ S = \frac{1}{2} \sum_{n=0}^{\infty} \frac{n(n+1)}{3^n} \]

### Step 3: Use the formula for the sum of a series
To find \( \sum_{n=0}^{\infty} n(n+1)x^n \), we can use the known result:
\[ \sum_{n=0}^{\infty} n(n+1)x^n = x \frac{d^2}{dx^2} \left( \frac{1}{1-x} \right) \]
Calculating this, we have:
\[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \]
Differentiating twice:
1. First derivative:
\[ \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2} \]
2. Second derivative:
\[ \frac{d^2}{dx^2} \left( \frac{1}{1-x} \right) = \frac{2}{(1-x)^3} \]
Thus,
\[ \sum_{n=0}^{\infty} n(n+1)x^n = x \cdot \frac{2}{(1-x)^3} \]

### Step 4: Substitute \( x = \frac{1}{3} \)
Now, substituting \( x = \frac{1}{3} \):
\[ \sum_{n=0}^{\infty} n(n+1) \left( \frac{1}{3} \right)^n = \frac{1}{3} \cdot \frac{2}{\left(1 - \frac{1}{3}\right)^3} = \frac{1}{3} \cdot \frac{2}{\left(\frac{2}{3}\right)^3} = \frac{1}{3} \cdot \frac{2 \cdot 27}{8} = \frac{9}{4} \]

### Step 5: Calculate \( S \)
Now substituting back into our expression for \( S \):
\[ S = \frac{1}{2} \cdot \frac{9}{4} = \frac{9}{8} \]

### Final Answer
Thus, the sum to infinity of the series is:
\[ \boxed{\frac{9}{8}} \]