If `(a^(n + 1) + b^(n + 1))/(a^(n) + b^(n))` is arithmetic mean of a and b, then n is equal to

Options

A. 1
B. 0
C. -1
D. none

Answer

To solve the problem, we need to find the value of \( n \) such that

\[ \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \]

is the arithmetic mean of \( a \) and \( b \). The arithmetic mean of \( a \) and \( b \) is given by

\[ \frac{a + b}{2}. \]

### Step-by-Step Solution:

1. **Set up the equation**: We start by equating the expression to the arithmetic mean:

\[ \frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \frac{a + b}{2}. \]

2. **Cross-multiply**: To eliminate the fraction, we cross-multiply:

\[ 2(a^{n+1} + b^{n+1}) = (a + b)(a^n + b^n). \]

3. **Expand the right side**: Now we expand the right-hand side:

\[ 2(a^{n+1} + b^{n+1}) = a^{n+1} + ab^n + b^{n+1} + ba^n. \]

4. **Rearrange the equation**: We can rearrange the equation to group similar terms:

\[ 2a^{n+1} + 2b^{n+1} = a^{n+1} + b^{n+1} + ab^n + ba^n. \]

This simplifies to:

\[ a^{n+1} + b^{n+1} = ab^n + ba^n. \]

5. **Factor out common terms**: Notice that we can factor the right-hand side:

\[ a^{n+1} + b^{n+1} = ab(a^{n-1} + b^{n-1}). \]

6. **Rearranging gives us**:

\[ a^{n+1} + b^{n+1} - ab(a^{n-1} + b^{n-1}) = 0. \]

7. **Set up a polynomial equation**: This can be rearranged into a polynomial form:

\[ a^{n+1} - aba^{n-1} + b^{n+1} - abb^{n-1} = 0. \]

8. **Factor out common terms**: We can factor this equation as:

\[ (a^{n} - b^{n})(a - b) = 0. \]

9. **Analyze the factors**: The equation \( (a^{n} - b^{n})(a - b) = 0 \) gives us two cases:
- \( a - b = 0 \) (which is not useful since we want \( a \neq b \))
- \( a^{n} - b^{n} = 0 \)

10. **Solve for \( n \)**: The equation \( a^{n} = b^{n} \) holds true if \( \frac{a}{b} = 1 \) or \( n = 0 \).

Thus, the only solution for \( n \) that satisfies the original equation is:

\[ \boxed{0}. \]