Sum of series `1+2/3+7/3^2+12/3^3 + . . . . . oo`

Answer

To find the sum of the series \( S = 1 + \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \ldots \), we can follow these steps:

### Step 1: Separate the first term
We can write the series as:
\[ S = 1 + \left(\frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \ldots\right) \]
Let \( T = \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \ldots \). Thus, we have:
\[ S = 1 + T \]

**Hint:** Isolate the first term to simplify the remaining series.

### Step 2: Analyze the remaining series
Now, we will analyze the series \( T \):
\[ T = \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \ldots \]

### Step 3: Identify a pattern
To find a pattern, let’s look at the numerators:
- The first term is \( 2 \)
- The second term is \( 7 \)
- The third term is \( 12 \)

The differences between the numerators are:
- \( 7 - 2 = 5 \)
- \( 12 - 7 = 5 \)

This suggests that the numerators can be expressed as:
- \( 2, 7, 12 \) can be represented as \( 2 + 5(n-1) \) for \( n = 1, 2, 3, \ldots \)

### Step 4: Rewrite the series
Thus, we can express \( T \) as:
\[ T = \sum_{n=1}^{\infty} \frac{2 + 5(n-1)}{3^n} \]
This can be split into two separate sums:
\[ T = \sum_{n=1}^{\infty} \frac{2}{3^n} + 5\sum_{n=1}^{\infty} \frac{(n-1)}{3^n} \]

### Step 5: Calculate the first sum
The first sum is a geometric series:
\[ \sum_{n=1}^{\infty} \frac{2}{3^n} = 2 \cdot \frac{1/3}{1 - 1/3} = 2 \cdot \frac{1/3}{2/3} = 1 \]

**Hint:** Use the formula for the sum of a geometric series.

### Step 6: Calculate the second sum
For the second sum, we use the formula for the sum of an infinite series involving \( n \):
\[ \sum_{n=0}^{\infty} n x^n = \frac{x}{(1-x)^2} \]
Thus, we can write:
\[ \sum_{n=1}^{\infty} \frac{(n-1)}{3^n} = \sum_{n=0}^{\infty} \frac{n}{3^{n+1}} = \frac{1/3}{(1 - 1/3)^2} = \frac{1/3}{(2/3)^2} = \frac{1/3}{4/9} = \frac{3}{4} \]

### Step 7: Combine the sums
Now substituting back, we have:
\[ T = 1 + 5 \cdot \frac{3}{4} = 1 + \frac{15}{4} = \frac{4}{4} + \frac{15}{4} = \frac{19}{4} \]

### Step 8: Find the total sum
Now we can find \( S \):
\[ S = 1 + T = 1 + \frac{19}{4} = \frac{4}{4} + \frac{19}{4} = \frac{23}{4} \]

### Final Answer
Thus, the sum of the series is:
\[ \boxed{\frac{23}{4}} \]