Find the sum of the series
`1^(2)+3^(2)+5^(2)+... ` to n terms

Answer

To find the sum of the series \(1^2 + 3^2 + 5^2 + \ldots\) up to \(n\) terms, we can follow these steps:

### Step 1: Identify the \(n\)th term
The series consists of the squares of the first \(n\) odd numbers. The \(n\)th odd number can be expressed as:
\[ t_n = 2n - 1 \]
Thus, the \(n\)th term of the series is:
\[ t_n^2 = (2n - 1)^2 = 4n^2 - 4n + 1 \]

### Step 2: Write the sum of the series
We need to find the sum of the first \(n\) terms:
\[ S_n = \sum_{k=1}^{n} t_k^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) \]

### Step 3: Separate the summation
We can separate the summation into three parts:
\[ S_n = 4\sum_{k=1}^{n} k^2 - 4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

### Step 4: Use the formulas for summation
We know the following formulas:
- The sum of the first \(n\) natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
- The sum of the squares of the first \(n\) natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
- The sum of \(1\) for \(n\) terms is simply \(n\):
\[ \sum_{k=1}^{n} 1 = n \]

### Step 5: Substitute the formulas into the summation
Now substituting these formulas into our expression for \(S_n\):
\[ S_n = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} - 4 \cdot \frac{n(n + 1)}{2} + n \]

### Step 6: Simplify the expression
Now simplifying each term:
1. The first term:
\[ 4 \cdot \frac{n(n + 1)(2n + 1)}{6} = \frac{2n(n + 1)(2n + 1)}{3} \]
2. The second term:
\[ -4 \cdot \frac{n(n + 1)}{2} = -2n(n + 1) \]
3. The third term remains \(n\).

Combining these, we have:
\[ S_n = \frac{2n(n + 1)(2n + 1)}{3} - 2n(n + 1) + n \]

### Step 7: Further simplification
Now we can factor out \(n(n + 1)\):
\[ S_n = n(n + 1) \left(\frac{2(2n + 1)}{3} - 2\right) + n \]
\[ = n(n + 1) \left(\frac{2(2n + 1) - 6}{3}\right) + n \]
\[ = n(n + 1) \left(\frac{4n - 4}{3}\right) + n \]
\[ = \frac{4n(n + 1)(n - 1)}{3} + n \]

### Step 8: Final expression
Thus, the final expression for the sum of the series \(S_n\) is:
\[ S_n = \frac{n}{3} (4n^2 - 1) = \frac{n(2n - 1)(2n + 1)}{3} \]