Find the sum of the series `1^2+3^2+5^2+ ton` terms.

Options

A. `(n(2n-1)(2n+1))/(3)`
B. `(n(2n+1)(2n+1))/(3)`
C. `(n(2n-1)(2n-1))/(3)`
D. `(n(2n+1)(2n-1))/(3)`

Explanation

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 series
The series consists of the squares of the first \(n\) odd numbers. The \(n\)-th odd number can be expressed as \(2n - 1\). Therefore, the series can be rewritten as:
\[ \sum_{k=1}^{n} (2k - 1)^2 \]

### Step 2: Expand the square
Now, we expand the square:
\[ (2k - 1)^2 = 4k^2 - 4k + 1 \]
Thus, the sum can be rewritten as:
\[ \sum_{k=1}^{n} (2k - 1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) \]

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

### Step 4: Use summation formulas
We will use the following formulas for summation:
- 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:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Substitute the formulas
Substituting these formulas into our expression gives:
\[ 4\left(\frac{n(n + 1)(2n + 1)}{6}\right) - 4\left(\frac{n(n + 1)}{2}\right) + n \]

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

Putting it all together:
\[ \frac{2n(n + 1)(2n + 1)}{3} - 2n(n + 1) + n \]

### Step 7: Combine like terms
Now, we can combine the terms:
\[ = \frac{2n(n + 1)(2n + 1) - 6n(n + 1) + 3n}{3} \]
\[ = \frac{2n(n + 1)(2n + 1) - 6n(n + 1) + 3n}{3} \]

### Step 8: Factor the expression
This can be factored to give:
\[ = \frac{n}{3}(2n + 1)(2n - 1) \]

### Final Result
Thus, the sum of the series \(1^2 + 3^2 + 5^2 + \ldots\) up to \(n\) terms is:
\[ \frac{n}{3}(2n + 1)(2n - 1) \]