Question

Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be the geometric mean between $a$ and $b$.

Answer 1

Given, $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be that Geometric mean between $a$ & $b$

$GM=\sqrt{ab}$

$\sqrt{ab}=\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$

$\left(ab{)}^{\frac{1}{2}}\right[a^n+b^n]=a^{n+1}+b^{n+1}$

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

$a^{(\frac{1}{2}+n)}[b^{\frac{1}{2}}-a^{\frac{1}{2}}]=b^{(n+\frac{1}{2})}[b^{\frac{1}{2}}-a^{\frac{1}{2}}]$

$a^{(\frac{1}{2}+n)}=\frac{b^{(n+\frac{1}{2})}[b^{\frac{1}{2}}-a^{\frac{1}{2}}]}{b^{\frac{1}{2}}-a^{\frac{1}{2}}}$

$a^{(\frac{1}{2}+n)}=b^{(n+\frac{1}{2})}\Rightarrow {\left(\frac{a}{b}\right)}^{(\frac{1}{2}+n)}=1$

$⟹\frac{1}{2}+n=0$

$⟹n=\frac{-1}{2}$