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 \(2^2 + 4^2 + 6^2 + 8^2 + \ldots\) 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 \(k\)-th term of the series can be expressed as:
\[ (2k)^2 \]
where \(k\) ranges from 1 to \(n\). Therefore, the terms can be rewritten as:
\[ 2^2, 4^2, 6^2, 8^2, \ldots, (2n)^2 \]

### Step 2: Express the series in summation notation
The sum of the series can be expressed as:
\[ \sum_{k=1}^{n} (2k)^2 \]

### Step 3: Simplify the summation
We can factor out the constant \(4\) from the summation:
\[ \sum_{k=1}^{n} (2k)^2 = \sum_{k=1}^{n} 4k^2 = 4 \sum_{k=1}^{n} k^2 \]

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

### Step 5: Final expression for the sum
Now we can simplify the expression:
\[ \sum_{k=1}^{n} (2k)^2 = \frac{4n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3} \]

### Conclusion
Thus, the sum of the series \(2^2 + 4^2 + 6^2 + 8^2 + \ldots\) up to \(n\) terms is:
\[ \frac{2n(n+1)(2n+1)}{3} \]

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