Find the sum of the series
`2^(2)+4^(2)+6^(2)+... ` to n terms.

Answer

To find the sum of the series \(2^2 + 4^2 + 6^2 + \ldots\) up to \(n\) terms, we can follow these steps:

### Step 1: Identify the nth term of the series
The series consists of the squares of even numbers. The \(n\)th term can be expressed as:
\[ a_n = (2n)^2 = 4n^2 \]
This means that the first term is \(2^2\) when \(n=1\), the second term is \(4^2\) when \(n=2\), and so on.

### Step 2: Write the sum of the series
The sum of the first \(n\) terms of the series can be written as:
\[ S_n = a_1 + a_2 + a_3 + \ldots + a_n = 4(1^2 + 2^2 + 3^2 + \ldots + n^2) \]

### Step 3: Use the formula for the sum of squares
The formula for the sum of the squares of the first \(n\) natural numbers is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]
Substituting this into our expression for \(S_n\):
\[ S_n = 4 \cdot \frac{n(n+1)(2n+1)}{6} \]

### Step 4: Simplify the expression
Now, we can simplify the expression:
\[ S_n = \frac{4n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3} \]

### Final Result
Thus, the sum of the series \(2^2 + 4^2 + 6^2 + \ldots\) up to \(n\) terms is:
\[ S_n = \frac{2n(n+1)(2n+1)}{3} \]