Find the sum of the series:
`(2^(2)+4^(2)+6^(2)+8^(2)+ ..."to n terms")`

Explanation

To find the sum of the series \( S = 2^2 + 4^2 + 6^2 + 8^2 + \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Identify the pattern in the series
The series consists of the squares of even numbers. The \( n \)-th term of the series can be expressed as \( (2k)^2 \) where \( k \) ranges from 1 to \( n \). Thus, we can rewrite the series as:

\[ S = (2 \cdot 1)^2 + (2 \cdot 2)^2 + (2 \cdot 3)^2 + \ldots + (2 \cdot n)^2 \]

### Step 2: Factor out the common term
Since each term has a common factor of \( 2^2 = 4 \), we can factor this out:

\[ S = 4 \left( 1^2 + 2^2 + 3^2 + \ldots + n^2 \right) \]

### Step 3: Use the formula for the sum of squares
The sum of the squares of the first \( n \) natural numbers is given by the formula:

\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 4: Substitute the formula into our expression for \( S \)
Now, substituting this formula into our expression for \( S \):

\[ S = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} \]

### Step 5: Simplify the expression
We can simplify this expression:

\[ S = \frac{4n(n + 1)(2n + 1)}{6} = \frac{2n(n + 1)(2n + 1)}{3} \]

### Final Answer
Thus, the sum of the series \( S = 2^2 + 4^2 + 6^2 + \ldots \) up to \( n \) terms is:

\[ S = \frac{2n(n + 1)(2n + 1)}{3} \]

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