Question Number 3466 by Rasheed Soomro last updated on 13/Dec/15
$${Find}\:{the}\:{value}\:{of}\:{n}\:{so}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} } \\ $$$${may}\:{become}\:\:{the}\:{A}.{M}.\:{between} \\ $$$${a}\:\:{and}\:\:{b}. \\ $$
Answered by prakash jain last updated on 13/Dec/15
$${n}=\mathrm{0}\:\mathrm{is}\:\mathrm{solution}. \\ $$$$\frac{{a}+{b}}{\mathrm{1}+\mathrm{1}}=\frac{{a}+{b}}{\mathrm{2}} \\ $$
Answered by RasheedSindhi last updated on 14/Dec/15
![((a^(n+1) +b^(n+1) )/(a^n +b^n ))=((a+b)/2) 2(a^(n+1) +b^(n+1) )=(a+b)(a^n +b^n =a^(n+1) +ab^n +a^n b+b^(n+1) a^(n+1) +b^(n+1) =ab^n +a^n b a^(n+1) −a^n b+b^(n+1) −ab^n =0 a^n (a−b)−b^n (a−b)=0 (a−b)(a^n −b^n )=0 a−b=0 ∣ a^n −b^n =0 •a=b ⇒ n may be any number in this case. •a^n =b^n ⇒(a^n /b^n )=1 [if b≠0] ((a/b))^n =((a/b))^0 ⇒n=0 [if b≠0]](https://www.tinkutara.com/question/Q3500.png)
$$ \\ $$$$\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }=\frac{{a}+{b}}{\mathrm{2}} \\ $$$$\mathrm{2}\left({a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} \right)=\left({a}+{b}\right)\left({a}^{{n}} +{b}^{{n}} \right. \\ $$$$\:\:\:\:\:\:\:\:\:\:={a}^{{n}+\mathrm{1}} +{ab}^{{n}} +{a}^{{n}} {b}+{b}^{{n}+\mathrm{1}} \\ $$$${a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} ={ab}^{{n}} +{a}^{{n}} {b} \\ $$$${a}^{{n}+\mathrm{1}} −{a}^{{n}} {b}+{b}^{{n}+\mathrm{1}} −{ab}^{{n}} =\mathrm{0} \\ $$$${a}^{{n}} \left({a}−{b}\right)−{b}^{{n}} \left({a}−{b}\right)=\mathrm{0} \\ $$$$\left({a}−{b}\right)\left({a}^{{n}} −{b}^{{n}} \right)=\mathrm{0} \\ $$$${a}−{b}=\mathrm{0}\:\mid\:{a}^{{n}} −{b}^{{n}} =\mathrm{0} \\ $$$$\bullet{a}={b}\:\:\Rightarrow\:{n}\:{may}\:{be}\:{any}\:{number} \\ $$$${in}\:{this}\:{case}. \\ $$$$\bullet{a}^{{n}} ={b}^{{n}} \Rightarrow\frac{{a}^{{n}} }{{b}^{{n}} }=\mathrm{1}\:\:\left[{if}\:{b}\neq\mathrm{0}\right] \\ $$$$\left(\frac{{a}}{{b}}\right)^{{n}} =\left(\frac{{a}}{{b}}\right)^{\mathrm{0}} \Rightarrow{n}=\mathrm{0}\:\:\:\left[{if}\:{b}\neq\mathrm{0}\right] \\ $$