Find the value of `n` so that `(a^(n+1)+b^(n+1))/(a^n+b^n)` may be the geometric mean between `a` and `b`.

Explanation

` therefore (a^(n+1)+b^(n+1))/(a^(n)+b^(n))=sqrt(ab)`
` implies (b^(n+1)[((a)/(b))^(n+1)+1])/(b^(n)[((a)/(b))^(n)+1])=bsqrt((a)/(b)) implies (((a)/(b))^(n+1)+1)/(((a)/(b))^(n)+1)=((a)/(b))^((1)/(2))`
Let `(a)/(b)= lambda`
` implies (lambda^(n+1)+1)/(lambda^(n)+1)=lambda^((1)/(2)) implies lambda^(n+1)+1=lambda^(n+(1)/(2))+lambda^((1)/(2)`
` implies lambda^(n+(1)/(2))(lambda^((1)/(2))-1)-(lambda^((1)/(2))-1)=0`
` implies (lambda^((1)/(2))-1)(lambda^(n+(1)/(2))-1)=0`
` implies lambda^((1)/(2))-1 ne 0 " " [ therefore a ne b]`
` therefore lambda^(n+(1)/(2))-1=0`
`implies lambda^(n+(1)/(2))=1=lambda^(0)`
`implies n+(1)/(2)=0 " or "n=-(1)/(2)`