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 $b$

Question

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 $b$.

My approach:

We know geometric mean between any two numbers $a$ and $b$ is given by $\sqrt{ab}$. By some processing in my brain I tried $n=-\frac{1}{2}$ and the result matched with the formula to find the geometric mean.

I wish to know is there any other disciplined approach for this problem?

Kindly guide me in this regard.

Answer 1

Suppose $a>b$

$\frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \sqrt{ab}\\ \implies a^{n+1} + b^{n+1}= a^{n+1/2}b^{1/2} + a^{1/2}b^{n+1/2} \\\implies a^{n+1/2}(a^{1/2}-b^{1/2}) - b^{n+1/2}(a^{1/2}-b^{1/2})=0\\ \implies (a^{1/2}-b^{1/2})(a^{n+1/2}-b^{n+1/2})=0 $

Since $a\ne b$, above equation forces $a^{n+1/2} = b^{n+1/2} \implies n+1/2=0$

Answer 2

Notice the following:

\begin{align*} \forall a,b\in\mathbb{R}^+\,\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{ab} &\iff \forall a,b\in\mathbb{R}^+a^{n+1}+b^{n+1}=a^n\sqrt{ab}+b^n\sqrt{ab}\\ &\iff \forall a,b\in\mathbb{R}^+ a^n\sqrt{a}(\sqrt{a}-\sqrt{b})+b^n\sqrt{b}(\sqrt{b}-\sqrt{a})=0\\ &\iff \forall a,b\in\mathbb{R}^+a^n\sqrt{a}(\sqrt{a}-\sqrt{b})=b^n\sqrt{b}(\sqrt{a}-\sqrt{b}) \end{align*}

Answer 3

Hint

For $a,b>0$ and $a\ne b$, the function$${a^{n+1}+b^{n+1}\over a^{n}+b^{n}}={a+b u^n\over 1+u^n}$$is absolutely increasing where $u={b\over a}$. There the answer $n=-{1\over 2}$ is the only such solution.

Answer 4

Let $a=2b$ in $$\frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \sqrt{ab}$$

and simplify you get $$\frac {2^{n+1 }+1}{2^{n}+1} = \sqrt 2$$

$$ 2^{n+1}+1=2^{1/2}+2^{n+\frac {1}{2}} $$

The only solution is $n=-1/2$

Answer 5

$\left(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}\right)$ is G.M. between $a$ and $b$.

$\Leftrightarrow \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=a^{(1 / 2)} b^{(1 / 2)} \quad\left[G . M\right.$. between $\left.a \& b=a^{(1 / 2)} b^{(1 / 2)}\right]$ $\Leftrightarrow a^{n+1}+b^{n+1}=a^{n+(1 / 2)} b^{(1 / 2)}+a^{(1 / 2)} b^{n+(1 / 2)}$ $\Leftrightarrow a^{n+1}-a^{n+(1 / 2)} b^{(1 / 2)}=a^{(1 / 2)} b^{n+(1 / 2)}-b^{n+1}$ $\Leftrightarrow a^{n+(1 / 2)}\left[a^{(1 / 2)}-b^{(1 / 2)}\right]=b^{n+(1 / 2)}\left[a^{(1 / 2)}-b^{(1 / 2)}\right]$ $\Leftrightarrow a^{n+(1 / 2)}=b^{n+(1 / 2)} \quad\left[\because a^{(1 / 2)}-b^{(1 / 2)} \neq 0\right.$, since $\left.a \neq b\right]$ $\Leftrightarrow\left(\frac{a}{b}\right)^{n+(1 / 2)}=1=\left(\frac{a}{b}\right)^{0}$ $\Leftrightarrow n+\frac{1}{2}=0$ $\Leftrightarrow n=-\frac{1}{2}$