`1^k + 2^k + 3^k + … + n^k` is divisble by 1 + 2 +3 + … n for every n` in N`, then K is :

Options

A. even
B. odd
C. only prime
D. none of these

Answer

To solve the problem, we need to determine the value of \( k \) such that the expression \( 1^k + 2^k + 3^k + \ldots + n^k \) is divisible by \( 1 + 2 + 3 + \ldots + n \) for every natural number \( n \).

### Step-by-Step Solution:

1. **Understanding the Sums**:
The sum \( 1 + 2 + 3 + \ldots + n \) can be calculated using the formula:
\[ S_n = \frac{n(n + 1)}{2} \]
This represents the sum of the first \( n \) natural numbers.

2. **Testing Values of \( n \)**:
We will test specific values of \( n \) to find a pattern for \( k \).

3. **Testing \( n = 1 \)**:
\[ 1^k \text{ is divisible by } 1 \]
This is true for any \( k \).

4. **Testing \( n = 2 \)**:
\[ 1^k + 2^k = 1 + 2^k \]
The sum \( S_2 = 1 + 2 = 3 \).
We need \( 1 + 2^k \) to be divisible by 3.
- For \( k = 0 \): \( 1 + 1 = 2 \) (not divisible)
- For \( k = 1 \): \( 1 + 2 = 3 \) (divisible)
- For \( k = 2 \): \( 1 + 4 = 5 \) (not divisible)
- For \( k = 3 \): \( 1 + 8 = 9 \) (divisible)

5. **Testing \( n = 3 \)**:
\[ 1^k + 2^k + 3^k = 1 + 2^k + 3^k \]
The sum \( S_3 = 1 + 2 + 3 = 6 \).
We need \( 1 + 2^k + 3^k \) to be divisible by 6.
- For \( k = 1 \): \( 1 + 2 + 3 = 6 \) (divisible)
- For \( k = 2 \): \( 1 + 4 + 9 = 14 \) (not divisible)
- For \( k = 3 \): \( 1 + 8 + 27 = 36 \) (divisible)

6. **Testing \( n = 4 \)**:
\[ 1^k + 2^k + 3^k + 4^k \]
The sum \( S_4 = 1 + 2 + 3 + 4 = 10 \).
We need \( 1 + 2^k + 3^k + 4^k \) to be divisible by 10.
- For \( k = 1 \): \( 1 + 2 + 3 + 4 = 10 \) (divisible)
- For \( k = 2 \): \( 1 + 4 + 9 + 16 = 30 \) (divisible)
- For \( k = 3 \): \( 1 + 8 + 27 + 64 = 100 \) (divisible)

7. **Conclusion**:
From the tests, we observe that \( k \) must be an odd number for the sums to be divisible by \( S_n \) for all tested values of \( n \).

Thus, the value of \( k \) is:
\[ \boxed{1} \text{ or any odd integer} \]