Given that ` 1^(2) + 2^(2) + 3^(2) + ... + 20^(2) = 2870`, the value of ` (2^(2) + 4^(2) + 6^(2) + ... + 40^(2))` is :

Options

A. 11480
B. 5740
C. 28700
D. 2870

Answer

To solve the problem, we need to find the value of the sum \(2^2 + 4^2 + 6^2 + ... + 40^2\).

### Step-by-step Solution:

1. **Identify the Series**:
The series \(2^2 + 4^2 + 6^2 + ... + 40^2\) can be rewritten in terms of \(n\):
\[ 2^2 + 4^2 + 6^2 + ... + 40^2 = (2 \cdot 1)^2 + (2 \cdot 2)^2 + (2 \cdot 3)^2 + ... + (2 \cdot 20)^2 \]
This means we are summing the squares of the first 20 even numbers.

2. **Factor Out the Common Term**:
We can factor out \(2^2\) from the series:
\[ = 2^2 \cdot (1^2 + 2^2 + 3^2 + ... + 20^2) \]

3. **Use the Given Information**:
We know from the problem statement that:
\[ 1^2 + 2^2 + 3^2 + ... + 20^2 = 2870 \]
So we can substitute this value into our equation:
\[ = 4 \cdot 2870 \]

4. **Calculate the Final Value**:
Now, we perform the multiplication:
\[ 4 \cdot 2870 = 11480 \]

Thus, the value of \(2^2 + 4^2 + 6^2 + ... + 40^2\) is **11480**.