If (aⁿ + bⁿ)/(aⁿ-1) + bn-1 is the A.M. between a and b, find n.

The A.M. between $a$ and $b$ is $\frac{a+b}{2}$.

$$\frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a+b}{2}$$

$$2a^n + 2b^n = a \cdot a^{n-1} + a \cdot b^{n-1} + b \cdot a^{n-1} + b \cdot b^{n-1}$$

Using the exponent rule $x^p \cdot x^q = x^{p+q}$:

$$2a^n + 2b^n = a^n + ab^{n-1} + ba^{n-1} + b^n$$

Subtract $a^n$ and $b^n$ from both sides:

$$2a^n - a^n + 2b^n - b^n = ab^{n-1} + ba^{n-1}$$ $$a^n + b^n = ab^{n-1} + ba^{n-1}$$

Move terms involving $a$ to one side and terms involving $b$ to the other side:

$$a^n - ba^{n-1} = ab^{n-1} - b^n$$

Factor $a^{n-1}$ from the left side and $b^{n-1}$ from the right side:

$$a^{n-1}(a - b) = b^{n-1}(a - b)$$

Assuming $a \neq b$, we can divide both sides by $(a-b)$:

$$a^{n-1} = b^{n-1}$$

Assuming $b \neq 0$, divide both sides by $b^{n-1}$:

$$\frac{a^{n-1}}{b^{n-1}} = 1$$

This can be written as:

$$\left(\frac{a}{b}\right)^{n-1} = 1$$

For this equation to hold, the exponent must be zero (assuming $\frac{a}{b} \neq 0, 1, -1$):

$$n-1 = 0$$ $$n = 1$$

The value of $n$ is $\boxed{1}$.