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

Explanation

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

### Step 1: Identify the series
The series can be expressed as:
\[ S = \sum_{n=0}^{\infty} \frac{(2n+1)^2}{2^{n+1}} \]
where \( (2n+1) \) represents the odd numbers.

### Step 2: Rewrite the series
We can rewrite the series as:
\[ S = \frac{1^2}{2^1} + \frac{3^2}{2^2} + \frac{5^2}{2^3} + \frac{7^2}{2^4} + \ldots \]
This can also be expressed as:
\[ S = \sum_{n=0}^{\infty} \frac{(2n+1)^2}{2^{n+1}} = \sum_{n=0}^{\infty} \frac{(4n^2 + 4n + 1)}{2^{n+1}} \]

### Step 3: Split the series
We can split the series into three parts:
\[ S = \sum_{n=0}^{\infty} \frac{4n^2}{2^{n+1}} + \sum_{n=0}^{\infty} \frac{4n}{2^{n+1}} + \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} \]

### Step 4: Calculate each part
1. **For the third part**:
\[ \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} = \frac{1/2}{1 - 1/2} = 1 \]

2. **For the second part** (using the formula for the sum of an arithmetic series):
\[ \sum_{n=0}^{\infty} \frac{n}{2^{n+1}} = \frac{1/2}{(1 - 1/2)^2} = 2 \]
Therefore,
\[ 4 \cdot \sum_{n=0}^{\infty} \frac{n}{2^{n+1}} = 4 \cdot 2 = 8 \]

3. **For the first part** (using the formula for the sum of squares):
\[ \sum_{n=0}^{\infty} \frac{n^2}{2^{n+1}} = \frac{2}{(1 - 1/2)^3} = 8 \]
Therefore,
\[ 4 \cdot \sum_{n=0}^{\infty} \frac{n^2}{2^{n+1}} = 4 \cdot 8 = 32 \]

### Step 5: Combine all parts
Now we can combine all parts:
\[ S = 32 + 8 + 1 = 41 \]

### Step 6: Final result
Thus, the sum of the series is:
\[ \boxed{41} \]

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