Question

For what value of n, $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is the arithmetic mean of a and b?

Answer 1

Since, arithmetic mean of a and b is $\frac{a+b}{2}$, therefore according to the given condition,

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

$\Rightarrow 2a^{n+1}+2b^{n+1}=a^{n+1}+a^{n}b+a{b}^n+b^{n+1}$

$\Rightarrow 2a^{n+1}-a^{n+1}-a^{n}b=a{b}^n+b^{n+1}-2b^{n+1}$

$\Rightarrow a^{n+1}-a^{n}b=a{b}^n-b^{n+1}$

$\Rightarrow a^n(a-b)=b^n(a-b)$

$\Rightarrow a^n=b^n[\because a\ne b]$

$\Rightarrow {\left(\frac{a}{b}\right)}^n=1\Rightarrow {\left(\frac{a}{b}\right)}^n={\left(\frac{a}{b}\right)}^0\Rightarrow n=0$

Hence, for n = 0, $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is the arithmetic mean of a and b.