Question Number 28320 by NECx last updated on 24/Jan/18
$${If}\:{the}\:{arithmetic}\:{mean}\:{of}\:{a}\:{and} \\ $$$${b}\:{is}\:\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{\mathrm{2}},\:{show}\:{that}\:{n}=\mathrm{0} \\ $$
Commented by mrW2 last updated on 24/Jan/18
$${this}\:{is}\:{not}\:{always}\:{true},\:{sir}. \\ $$$${e}.{g}.\:{if}\:{a}={b}=\mathrm{1},\:{or}\:{if}\:{a}=\mathrm{1}\:{and}\:{b}=\mathrm{0},\:{n} \\ $$$${can}\:{be}\:{any}\:{value}. \\ $$
Commented by NECx last updated on 24/Jan/18
$${thanks}\:{for}\:{the}\:{correction} \\ $$
Commented by abdo imad last updated on 24/Jan/18
$$\frac{{a}^{{n}+\mathrm{1}} \:+{b}^{{n}+\mathrm{1}} }{\mathrm{2}}\:=\:\frac{{a}+{b}}{\mathrm{2}}\:\Leftrightarrow\:{a}^{{n}+\mathrm{1}} −{a}\:={b}^{{n}+\mathrm{1}} −{b}\:\forall{a}\:{andb}\: \\ $$$$\Rightarrow\frac{\partial}{\partial{a}}\left(\:{a}^{{n}+\mathrm{1}} −{a}\right)=\mathrm{0}\:{for}\:{bfixed}\:\Rightarrow\left({n}+\mathrm{1}\right){a}^{{n}} −\mathrm{1}=\mathrm{0}\:\:\forall{a} \\ $$$$\Rightarrow\left({n}+\mathrm{1}\right){a}^{{n}} \:=\mathrm{1}\:\:\:\forall{a}\:\Rightarrow{n}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$