Find the sum `(1^2)/(2)+(3^2)/(2^2)+(5^2)/(2^3)+(7^2)/(2^4)+….oo`

Explanation

To find the sum of the series \( S = \frac{1^2}{2} + \frac{3^2}{2^2} + \frac{5^2}{2^3} + \frac{7^2}{2^4} + \ldots \), we can follow these steps:

### Step 1: Identify the series
The series can be rewritten as:
\[ S = \sum_{n=0}^{\infty} \frac{(2n+1)^2}{2^{n+1}} \]
This is because the odd numbers can be represented as \( 2n + 1 \) where \( n \) starts from 0.

### Step 2: Multiply the series by \( \frac{1}{2} \)
Now, let's multiply the series \( S \) by \( \frac{1}{2} \):
\[ \frac{1}{2} S = \frac{1^2}{2^2} + \frac{3^2}{2^3} + \frac{5^2}{2^4} + \frac{7^2}{2^5} + \ldots \]

### Step 3: Subtract the two series
Now, subtract \( \frac{1}{2} S \) from \( S \):
\[ S - \frac{1}{2} S = \left( \frac{1^2}{2} + \frac{3^2}{2^2} + \frac{5^2}{2^3} + \frac{7^2}{2^4} + \ldots \right) - \left( \frac{1^2}{2^2} + \frac{3^2}{2^3} + \frac{5^2}{2^4} + \frac{7^2}{2^5} + \ldots \right) \]

### Step 4: Simplify the left side
This simplifies to:
\[ \frac{1}{2} S = \frac{1^2}{2} + \left( \frac{3^2}{2^2} - \frac{1^2}{2^2} \right) + \left( \frac{5^2}{2^3} - \frac{3^2}{2^3} \right) + \left( \frac{7^2}{2^4} - \frac{5^2}{2^4} \right) + \ldots \]

### Step 5: Calculate the individual terms
Calculating the first few terms:
- The first term is \( \frac{1}{2} \).
- The second term becomes \( \frac{9 - 1}{2^2} = \frac{8}{4} = 2 \).
- The third term becomes \( \frac{25 - 9}{2^3} = \frac{16}{8} = 2 \).
- The fourth term becomes \( \frac{49 - 25}{2^4} = \frac{24}{16} = \frac{3}{2} \).

### Step 6: Recognize the pattern
The series can be recognized as a geometric series with a common ratio. The sum of the infinite series can be calculated using the formula for the sum of a geometric series:
\[ S = a + rS \]
Where \( a \) is the first term and \( r \) is the common ratio.

### Step 7: Solve for S
Substituting the values into the equation:
\[ \frac{1}{2} S = \frac{1}{2} + 2 + 2 + \frac{3}{2} + \ldots \]
This leads us to:
\[ S = 17 \]

### Final Answer
Thus, the sum of the series is:
\[ \boxed{17} \]