If `1^3 + 2^3 +3^3 +.....10^3` = 3025 , then find the value of `2^3+4^3+6^3 + .....+ 20^3`

Options

A. 6050
B. 9075
C. 12100
D. 24200

Answer

To solve the problem, we need to find the value of \(2^3 + 4^3 + 6^3 + \ldots + 20^3\).

### Step-by-Step Solution:

1. **Identify the series**: The series \(2^3 + 4^3 + 6^3 + \ldots + 20^3\) consists of the cubes of the first 10 even numbers.

2. **Rewrite the series**: We can express the even numbers in terms of their position in the series:
\[ 2^3 + 4^3 + 6^3 + \ldots + 20^3 = (2 \cdot 1)^3 + (2 \cdot 2)^3 + (2 \cdot 3)^3 + \ldots + (2 \cdot 10)^3 \]

3. **Factor out the common term**: Since each term is multiplied by 2, we can factor out \(2^3\) from the series:
\[ = 2^3 \cdot (1^3 + 2^3 + 3^3 + \ldots + 10^3) \]

4. **Use the given value**: We know from the problem statement that \(1^3 + 2^3 + 3^3 + \ldots + 10^3 = 3025\). Therefore:
\[ = 2^3 \cdot 3025 \]

5. **Calculate \(2^3\)**: We know that \(2^3 = 8\).

6. **Multiply**: Now we can multiply:
\[ 8 \cdot 3025 = 24200 \]

Thus, the value of \(2^3 + 4^3 + 6^3 + \ldots + 20^3\) is **24200**.

### Final Answer:
\[ \boxed{24200} \]