Statement 1 The sum of the series `1+(1+2+4)+(4+6+9)+(9+12+16)+"……."+(361+380+400)` is 8000.
Statement 2 `sum_(k=1)^(n)(k^(3)-(k-1)^(3))=n^(3)` for any natural number n.

Options

A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1
B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1
C. Statement 1 is true, Statement 2 is false
D. Statement 1 is false, Statement 2 is true

Explanation

To solve the given problem, we need to analyze both statements separately.

### Statement 1:
The sum of the series \(1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \ldots + (361 + 380 + 400)\) is 8000.

1. **Identify the pattern in the series**:
- The first term is \(1\).
- The second term is \(1 + 2 + 4 = 7\).
- The third term is \(4 + 6 + 9 = 19\).
- The fourth term is \(9 + 12 + 16 = 37\).
- The last term is \(361 + 380 + 400\).

2. **Rewrite the terms using the difference of cubes**:
- Notice that each group can be expressed as the difference of cubes:
- \(1 = 1^3 - 0^3\)
- \(7 = 2^3 - 1^3\)
- \(19 = 3^3 - 2^3\)
- ...
- The last term can be expressed as \(20^3 - 19^3\).

3. **Generalize the series**:
- The series can be rewritten as:
\[ \sum_{k=1}^{20} (k^3 - (k-1)^3) \]

4. **Apply the telescoping property**:
- When we sum these differences, all intermediate terms cancel out:
\[ (1^3 - 0^3) + (2^3 - 1^3) + (3^3 - 2^3) + \ldots + (20^3 - 19^3) = 20^3 - 0^3 = 20^3 \]

5. **Calculate \(20^3\)**:
- \(20^3 = 8000\).

Thus, **Statement 1 is true**.

### Statement 2:
\(\sum_{k=1}^{n} (k^3 - (k-1)^3) = n^3\) for any natural number \(n\).

1. **Expand the left-hand side**:
- Using the identity for the difference of cubes:
\[ k^3 - (k-1)^3 = k^3 - (k^3 - 3k^2 + 3k - 1) = 3k^2 - 3k + 1 \]

2. **Sum the series**:
- Therefore, we have:
\[ \sum_{k=1}^{n} (k^3 - (k-1)^3) = \sum_{k=1}^{n} (3k^2 - 3k + 1) \]

3. **Separate the sums**:
- This can be separated into three sums:
\[ 3\sum_{k=1}^{n} k^2 - 3\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

4. **Use known formulas**:
- We know:
- \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)
- \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)
- \(\sum_{k=1}^{n} 1 = n\)

5. **Substitute and simplify**:
- Substitute these into the equation:
\[ 3 \cdot \frac{n(n+1)(2n+1)}{6} - 3 \cdot \frac{n(n+1)}{2} + n \]
- Simplifying this gives:
\[ \frac{n(n+1)(2n+1)}{2} - \frac{3n(n+1)}{2} + n \]
- Combine terms:
\[ \frac{n(n+1)(2n+1 - 3)}{2} + n = \frac{n(n+1)(2n - 2)}{2} + n = n^3 \]

Thus, **Statement 2 is also true**.

### Final Conclusion:
Both statements are true.