Find n, so that `(a^(n+1)+b^(n+1))/(a^(n)+b^(n))(a!=b)` be the HM between a and b.

Options

A. n=0
B. n=13
C. `n=-1//2`
D. `n=-1`

Answer

To find \( n \) such that

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

is the harmonic mean (HM) of \( a \) and \( b \), we start by recalling the formula for the harmonic mean of two numbers \( a \) and \( b \):

\[ HM = \frac{2ab}{a + b} \]

We want to set the two expressions equal to each other:

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

### Step 1: Cross-multiply

Cross-multiplying gives us:

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

### Step 2: Expand both sides

Expanding both sides, we get:

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

### Step 3: Rearranging the equation

Rearranging the equation, we have:

\[ a^{n+2} + b^{n+2} + a^{n+1}b + ab^{n+1} - 2a^{n+1}b - 2ab^{n+1} = 0 \]

This simplifies to:

\[ a^{n+2} + b^{n+2} - a^{n+1}b - ab^{n+1} = 0 \]

### Step 4: Factor the equation

We can factor this equation:

\[ a^{n+2} - a^{n+1}b + b^{n+2} - ab^{n+1} = 0 \]

This can be rearranged as:

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

### Step 5: Simplifying further

Factoring out \( (a - b) \):

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

Since \( a \neq b \), we can set:

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

### Step 6: Solving for \( n \)

This implies:

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

Taking logarithms gives:

\[ (n + 1) \log a = (n + 1) \log b \]

Since \( a \neq b \), we can conclude that:

\[ n + 1 = 0 \implies n = -1 \]

### Final Answer

Thus, the value of \( n \) is:

\[ \boxed{-1} \]