The sum of infinite terms of the series `5 - 7/3 + (9)/(3 ^(2)) - (11)/( 3 ^(3))+……..o` is

Answer

To find the sum of the infinite series \( S = 5 - \frac{7}{3} + \frac{9}{3^2} - \frac{11}{3^3} + \ldots \), we will follow these steps:

### Step 1: Identify the series
The series can be expressed as:
\[ S = 5 - \frac{7}{3} + \frac{9}{3^2} - \frac{11}{3^3} + \ldots \]
We can see that the numerators \( 5, 7, 9, 11, \ldots \) form an arithmetic progression (AP) with a common difference of 2.

### Step 2: Write the series in terms of \( n \)
The general term can be expressed as:
\[ a_n = \frac{(2n + 3)}{3^{n-1}} (-1)^{n-1} \]
where \( n = 1, 2, 3, \ldots \)

### Step 3: Multiply the series by 3
Now, we multiply the entire series \( S \) by \( 3 \):
\[ 3S = 15 - 7 + \frac{9}{3} - \frac{11}{3^2} + \ldots \]
This shifts the terms to the right.

### Step 4: Write the shifted series
The shifted series can be expressed as:
\[ 3S = 15 - 7 + 3 - \frac{11}{3^2} + \ldots \]

### Step 5: Combine the original and shifted series
Now, we will add \( S \) and \( 3S \):
\[ S + 3S = 4S = 15 - \frac{7}{3} + 3 - \frac{11}{3^2} + \ldots \]
\[ 4S = 15 - \frac{7}{3} + \left( \frac{9}{3^2} - \frac{11}{3^3} + \ldots \right) \]

### Step 6: Simplify the series
We can factor out the common terms:
\[ 4S = 15 - \frac{7}{3} + \frac{2}{3} \left( 1 - \frac{1}{3} + \frac{1}{3^2} - \ldots \right) \]
The series \( 1 - \frac{1}{3} + \frac{1}{3^2} - \ldots \) is a geometric series with first term \( 1 \) and common ratio \( -\frac{1}{3} \).

### Step 7: Calculate the sum of the geometric series
The sum of the infinite geometric series is given by:
\[ \text{Sum} = \frac{1}{1 - r} = \frac{1}{1 - (-\frac{1}{3})} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]
Thus, substituting this back:
\[ 4S = 15 - \frac{7}{3} + \frac{2}{3} \cdot \frac{3}{4} \]
\[ 4S = 15 - \frac{7}{3} + \frac{1}{2} \]

### Step 8: Combine the terms
Convert \( 15 \) and \( \frac{1}{2} \) to have a common denominator of \( 6 \):
\[ 15 = \frac{90}{6}, \quad \frac{1}{2} = \frac{3}{6} \]
Thus,
\[ 4S = \frac{90}{6} - \frac{14}{6} + \frac{3}{6} = \frac{90 - 14 + 3}{6} = \frac{79}{6} \]

### Step 9: Solve for \( S \)
Now, divide both sides by 4:
\[ S = \frac{79}{24} \]

### Final Answer
The sum of the infinite series is:
\[ S = \frac{79}{24} \]