`(x^(n)+y^(n))` is divisible by `(x+y)` is true when `n in NN` is of the form `(k in N)`:

Options

A. 4k+1
B. 4k+3
C. 4k+7
D. 2k

Answer

To determine the values of \( n \) for which \( (x^n + y^n) \) is divisible by \( (x + y) \), we can use the principle of mathematical induction and properties of polynomial divisibility.

### Step-by-Step Solution:

1. **Base Case (n = 1)**:
- For \( n = 1 \):
\[ x^1 + y^1 = x + y \]
- Clearly, \( x + y \) is divisible by \( x + y \). Thus, the base case holds true.

**Hint**: Check the simplest case first to establish a foundation for induction.

2. **Inductive Step**:
- Assume that for some odd integer \( n = 2k + 1 \), \( x^n + y^n \) is divisible by \( x + y \).
- We need to show that \( x^{n+2} + y^{n+2} \) is also divisible by \( x + y \).

3. **Express \( n + 2 \)**:
- Let \( n + 2 = 2k + 1 + 2 = 2k + 3 \).
- We need to show that \( x^{2k+3} + y^{2k+3} \) is divisible by \( x + y \).

4. **Using Polynomial Factorization**:
- We can express \( x^{2k+3} + y^{2k+3} \) as:
\[ x^{2k+3} + y^{2k+3} = x^{2k+1} \cdot x^2 + y^{2k+1} \cdot y^2 \]
- We can factor this as:
\[ (x + y)(\text{some polynomial in } x \text{ and } y) \]

5. **Verification**:
- To verify, we can add and subtract terms to manipulate the expression:
\[ x^{2k+3} + y^{2k+3} = (x^{2k+1} + y^{2k+1})(x + y) + \text{(some terms)} \]
- Since by the inductive hypothesis \( x^{2k+1} + y^{2k+1} \) is divisible by \( x + y \), the entire expression remains divisible by \( x + y \).

6. **Conclusion**:
- By the principle of mathematical induction, we conclude that \( x^n + y^n \) is divisible by \( x + y \) for all odd integers \( n \).

### Final Result:
Thus, \( n \) must be of the form \( n = 2k + 1 \) where \( k \in \mathbb{N} \).