Question

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

Or

If f is a function satisfying $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in N$ such that f(1)=3 and ${\sum }_{x=1}^{n}f\left(x\right)=120$ find the value of n.

Answer 1

It is given that $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$is the Gm between a and b.

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

$\Rightarrow a^{n=1}+b^{n+1}=a^{1/2}.b^{1/2}(a^n+b^n)$

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

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

$\Rightarrow a^{n+\frac{1}{2}}=b^{n+\frac{1}{2}}[\because a^{1/2}-b^{1/2}\ne 0,asa\ne b]$

$\Rightarrow \frac{a^{n+1/2}}{b^{n+1/2}}=1\Rightarrow {\left(\frac{a}{b}\right)}^{n+\frac{1}{2}}={\left(\frac{a}{b}\right)}^0$

$\Rightarrow n+\frac{1}{2}=0\Rightarrow =-\frac{1}{2}$

Or

we have, $f(x+y)=f\left(x\right).f\left(y\right)forx,y\in N$

Given ,$f\left(1\right)=3and{\sum }_{x=1}^n=120$

Now,${\sum }_{x=1}^{n}f\left(x\right)=f\left(1\right)+f\left(2\right)+f\left(3\right)+f\left(4\right)+.....+f\left(n\right)$

$\Rightarrow 120=f\left(1\right)+\underset{2times}{\underset{}{f\left(1\right).f\left(1\right)}}+\underset{3times}{\underset{}{f\left(1\right).f\left(1\right).f\left(1\right)}}+...+\underset{ntimes}{\underset{}{f\left(1\right).f\left(1\right)...f\left(1\right)}}$

$\Rightarrow 120=3+3^2+3^3+...+3^n$

$\Rightarrow 120\frac{3(3^n-1)}{3-1}$

$\Rightarrow 3^n-1=80\Rightarrow 3^n=81$

$\Rightarrow 3^n=3^4\Rightarrow n=4$