If the arthimetic mean between a and b is $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ , then n =

Given: $\frac{{\mathrm{a}}^{\mathrm{n}+1}+{\mathrm{b}}^{\mathrm{n}+1}}{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}$  is an arithmetic mean of a and b.

Therefore,

$\frac{{\mathrm{a}}^{\mathrm{n}+1}+{\mathrm{b}}^{\mathrm{n}+1}}{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}=\frac{\mathrm{a}+\mathrm{b}}{2}$   ...(i)

L.H.S $=\frac{{\mathrm{a}}^{\mathrm{n}+1}+{\mathrm{b}}^{\mathrm{n}+1}}{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}$

Putting n = 0, we get

$\frac{{\mathrm{a}}^{\mathrm{n}+1}+{\mathrm{b}}^{\mathrm{n}+1}}{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}=\frac{{\mathrm{a}}^{0+1}+{\mathrm{b}}^{0+1}}{{\mathrm{a}}^0+{\mathrm{b}}^0}$

$=\frac{\mathrm{a}+\mathrm{b}}{1+1}$   [∵ a⁰ = 1]

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

So, for n = 0, the given expression is an arithmetic mean of a and b.