The vlaue of the `Sigma_(n=0)^(oo) (2n+3)/(3^n)` is equal to ______.

Explanation

To solve the problem, we need to find the value of the infinite series given by:

\[ S = \sum_{n=0}^{\infty} \frac{2n + 3}{3^n} \]

We can break this series into two separate summations:

\[ S = \sum_{n=0}^{\infty} \frac{2n}{3^n} + \sum_{n=0}^{\infty} \frac{3}{3^n} \]

Let's denote the first summation as \( S_1 \) and the second summation as \( S_2 \).

### Step 1: Calculate \( S_2 \)

The second summation is:

\[ S_2 = \sum_{n=0}^{\infty} \frac{3}{3^n} = 3 \sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n \]

This is a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{3} \). The sum of an infinite geometric series is given by:

\[ \text{Sum} = \frac{a}{1 - r} \]

So, we have:

\[ S_2 = 3 \cdot \frac{1}{1 - \frac{1}{3}} = 3 \cdot \frac{1}{\frac{2}{3}} = 3 \cdot \frac{3}{2} = \frac{9}{2} \]

### Step 2: Calculate \( S_1 \)

Now we need to calculate \( S_1 \):

\[ S_1 = \sum_{n=0}^{\infty} \frac{2n}{3^n} = 2 \sum_{n=0}^{\infty} \frac{n}{3^n} \]

To find \( \sum_{n=0}^{\infty} \frac{n}{3^n} \), we can use the formula for the sum of \( n \cdot r^n \):

\[ \sum_{n=0}^{\infty} n r^n = \frac{r}{(1 - r)^2} \]

In our case, \( r = \frac{1}{3} \):

\[ \sum_{n=0}^{\infty} n \left(\frac{1}{3}\right)^n = \frac{\frac{1}{3}}{\left(1 - \frac{1}{3}\right)^2} = \frac{\frac{1}{3}}{\left(\frac{2}{3}\right)^2} = \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{9}{12} = \frac{3}{4} \]

Thus,

\[ S_1 = 2 \cdot \frac{3}{4} = \frac{3}{2} \]

### Step 3: Combine \( S_1 \) and \( S_2 \)

Now we can combine both results to find \( S \):

\[ S = S_1 + S_2 = \frac{3}{2} + \frac{9}{2} = \frac{12}{2} = 6 \]

Thus, the value of the series is:

\[ \boxed{6} \]