A Tricky , though easy... Chase ..!!

$\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ = $AM$ ( $a, b$ ) = $\frac{a+b}{2}$

$2a^{n+1}+2b^{n+1} = (a+b)(a^{n}+b^{n})$

$a^{n+1} - a^nb + b^{n+1} - ab^n = 0$

$a^n(a-b)+b^n(b-a)=0$ $(a^n-b^n)(a-b) = 0$

Either $(a^n-b^n)$ or $(a-b)$ is equal to $0$

Problem doesn't indicate that $a = b$ , so $a^n-b^n = 0$ $a^n = b^n$

Since we aren't given that $a = b$ , the only solution is when $n = 0$

$AM(a,b)=\frac{a+b}{2}$

The exponent of $a$ and $b$ is (and must be ) 1. $\frac{a^1+b^1}{2}$

$n$ must be 0 as only $(n=0)$ meets this condition ( $0+1=1$ ), in this case:

$\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$

Substitute values into the equation:

$\frac{a^{0+1}+b^{0+1}}{a^0+b^0}$

Resulting equation:

$\frac{a+b}{1+1}$ = $\frac{a+b}{2}$

Answer: $n=\boxed{0}$

AM of a and b is a+b/2.

So, n=0.