Require (\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\dfrac{2ab}{a+b})
Similar algebra as A.M. case leads to (n=-1).
Check: (\dfrac{a^{0}+b^{0}}{a^{-1}+b^{-1}}=\dfrac{2}{\frac{a+b}{ab}}=\dfrac{2ab}{a+b}) ✓
Answer: (n=-1)
Require (\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\dfrac{2ab}{a+b})
Similar algebra as A.M. case leads to (n=-1).
Check: (\dfrac{a^{0}+b^{0}}{a^{-1}+b^{-1}}=\dfrac{2}{\frac{a+b}{ab}}=\dfrac{2ab}{a+b}) ✓
Answer: (n=-1)