If `1^(2)+2^(2)+3^(2)+......+512^(2)=m`, then `2^(2)+4^(2)+6^(2)+.........+1024^(2)` is equal to

Options

A. 3m
B. 4m
C. `m^(2)`
D. `m^(3)`

Answer

To solve the problem, we need to find the value of the sum \( 2^2 + 4^2 + 6^2 + \ldots + 1024^2 \) given that \( 1^2 + 2^2 + 3^2 + \ldots + 512^2 = m \).

### Step-by-Step Solution:

1. **Understanding the Given Expression:**
We know that:
\[ 1^2 + 2^2 + 3^2 + \ldots + 512^2 = m \]

2. **Identifying the New Series:**
The series \( 2^2 + 4^2 + 6^2 + \ldots + 1024^2 \) can be rewritten in terms of the first series. Notice that:
\[ 2^2 + 4^2 + 6^2 + \ldots + 1024^2 = (2 \cdot 1)^2 + (2 \cdot 2)^2 + (2 \cdot 3)^2 + \ldots + (2 \cdot 512)^2 \]

3. **Factoring Out the Common Factor:**
We can factor out \( 2^2 \) from each term:
\[ = 2^2(1^2 + 2^2 + 3^2 + \ldots + 512^2) \]

4. **Substituting the Value of m:**
From the first step, we know that:
\[ 1^2 + 2^2 + 3^2 + \ldots + 512^2 = m \]
Therefore, we can substitute \( m \) into our equation:
\[ = 2^2 \cdot m \]

5. **Calculating the Final Result:**
Since \( 2^2 = 4 \), we have:
\[ = 4m \]

### Final Answer:
Thus, the value of \( 2^2 + 4^2 + 6^2 + \ldots + 1024^2 \) is:
\[ \boxed{4m} \]