For what value of `n , (a ^(n +3) + b ^(n +3))/(a ^(n +2) + b ^(n +2)), a ne b` in the A.M. of a and b.

Answer

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

\[ \frac{a^{(n+3)} + b^{(n+3)}}{a^{(n+2)} + b^{(n+2)}} \]

is equal to the arithmetic mean of \( a \) and \( b \), which is given by

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

### Step 1: Set up the equation

We start by equating the two expressions:

\[ \frac{a^{(n+3)} + b^{(n+3)}}{a^{(n+2)} + b^{(n+2)}} = \frac{a + b}{2} \]

### Step 2: Cross-multiply

Cross-multiplying gives us:

\[ 2(a^{(n+3)} + b^{(n+3)}) = (a + b)(a^{(n+2)} + b^{(n+2)}) \]

### Step 3: Expand the right-hand side

Expanding the right-hand side:

\[ 2(a^{(n+3)} + b^{(n+3)}) = a^{(n+3)} + ab^{(n+2)} + b^{(n+3)} + ba^{(n+2)} \]

### Step 4: Rearranging the equation

Now, we rearrange the equation:

\[ 2a^{(n+3)} + 2b^{(n+3)} = a^{(n+3)} + ab^{(n+2)} + b^{(n+3)} + ba^{(n+2)} \]

This simplifies to:

\[ a^{(n+3)} + b^{(n+3)} = ab^{(n+2)} + ba^{(n+2)} \]

### Step 5: Factor out common terms

Factoring out common terms, we can rewrite it as:

\[ a^{(n+3)} + b^{(n+3)} = ab(a^{(n+1)} + b^{(n+1)}) \]

### Step 6: Compare powers of \( a \) and \( b \)

Now we will compare the powers of \( a \) and \( b \). For the left-hand side, the highest power is \( n + 3 \), and for the right-hand side, the highest power is \( n + 1 \).

Setting these equal gives us:

1. From the left side: \( n + 3 \)
2. From the right side: \( n + 1 \)

Thus, we have:

\[ n + 3 = n + 1 \]

### Step 7: Solve for \( n \)

Subtract \( n \) from both sides:

\[ 3 = 1 \]

This is not possible, so we need to consider the other equation we derived from the denominator:

\[ n + 2 = 0 \]

Solving for \( n \):

\[ n = -2 \]

### Final Answer

Thus, the value of \( n \) for which the expression is the arithmetic mean of \( a \) and \( b \) is:

\[ \boxed{-2} \]