The sum of the series `1+ 5 ((4n +1)/(4n - 3)) + 9 ((4n +1) /( 4n -3)) ^(2) + 13 ((4n +1)/( 4n -3)) ^(3) + …….` is .

Answer

To find the sum of the series \( S = 1 + 5 \left( \frac{4n + 1}{4n - 3} \right) + 9 \left( \frac{4n + 1}{4n - 3} \right)^2 + 13 \left( \frac{4n + 1}{4n - 3} \right)^3 + \ldots \), we will follow these steps:

### Step 1: Define the variable
Let \( x = \frac{4n + 1}{4n - 3} \). Then we can rewrite the series as:
\[ S = 1 + 5x + 9x^2 + 13x^3 + \ldots \]

### Step 2: Identify the pattern
Notice that the coefficients \( 1, 5, 9, 13, \ldots \) form an arithmetic progression (AP) with the first term \( a = 1 \) and a common difference \( d = 4 \).

### Step 3: Write the series in terms of \( S \)
We can express \( S \) in a more manageable form:
\[ S = 1 + 5x + 9x^2 + 13x^3 + \ldots \]

### Step 4: Multiply the series by \( x \)
Now, we multiply the entire series \( S \) by \( x \):
\[ xS = 5x + 9x^2 + 13x^3 + \ldots \]

### Step 5: Subtract the two equations
Subtract the equation for \( xS \) from \( S \):
\[ S - xS = 1 + (5x - 5x) + (9x^2 - 5x^2) + (13x^3 - 9x^2) + \ldots \]
This simplifies to:
\[ S(1 - x) = 1 + 4x + 4x^2 + 4x^3 + \ldots \]

### Step 6: Recognize the geometric series
The series \( 4x + 4x^2 + 4x^3 + \ldots \) is a geometric series with first term \( 4x \) and common ratio \( x \):
\[ \text{Sum} = \frac{4x}{1 - x} \]

### Step 7: Substitute back into the equation
Now we have:
\[ S(1 - x) = 1 + \frac{4x}{1 - x} \]

### Step 8: Solve for \( S \)
Multiply both sides by \( (1 - x) \):
\[ S(1 - x) = 1 - x + 4x \]
\[ S(1 - x) = 1 + 3x \]
\[ S = \frac{1 + 3x}{1 - x} \]

### Step 9: Substitute \( x \) back
Now substitute \( x = \frac{4n + 1}{4n - 3} \):
\[ S = \frac{1 + 3\left(\frac{4n + 1}{4n - 3}\right)}{1 - \left(\frac{4n + 1}{4n - 3}\right)} \]

### Step 10: Simplify the expression
Calculating the numerator:
\[ 1 + 3\left(\frac{4n + 1}{4n - 3}\right) = \frac{(4n - 3) + 3(4n + 1)}{4n - 3} = \frac{4n - 3 + 12n + 3}{4n - 3} = \frac{16n}{4n - 3} \]

Calculating the denominator:
\[ 1 - \left(\frac{4n + 1}{4n - 3}\right) = \frac{(4n - 3) - (4n + 1)}{4n - 3} = \frac{-4}{4n - 3} \]

Thus:
\[ S = \frac{\frac{16n}{4n - 3}}{\frac{-4}{4n - 3}} = -4n \]

### Final Answer
The sum of the series is:
\[ S = 4n \]