Question Number 171861 by Mikenice last updated on 21/Jun/22
$${if}\: \\ $$$${ax}+{by}=\mathrm{5} \\ $$$${ax}^{\mathrm{2}} +{by}^{\mathrm{2}} =\mathrm{10} \\ $$$${ax}^{\mathrm{3}} +{by}^{\mathrm{3}} =\mathrm{50} \\ $$$${ax}^{\mathrm{4}} +{by}^{\mathrm{4}} =\mathrm{130} \\ $$$${find}\:\mathrm{13}\left({x}+{y}−{xy}\right)−\mathrm{120}\left({a}+{b}\right) \\ $$
Commented by infinityaction last updated on 21/Jun/22
$$\mathrm{42} \\ $$
Commented by Mikenice last updated on 21/Jun/22
$${please}\:{sir}\:{show}\:{the}\:{solution} \\ $$
Commented by mr W last updated on 21/Jun/22
$${it}'{s}\:{more}\:{interesting}\:{and} \\ $$$${challenging}\:{to}\:{find}\:{ax}^{\mathrm{5}} +{by}^{\mathrm{5}} ,\:{ax}^{\mathrm{6}} +{by}^{\mathrm{6}} \\ $$$${or}\:{generally}\:{ax}^{{n}} +{by}^{{n}} =? \\ $$
Commented by Mikenice last updated on 21/Jun/22
$${if}\:{you}\:{can}\:{show}\:{solution},\:{please}\:{do}\:{so} \\ $$
Commented by mr W last updated on 21/Jun/22
$$\left({ax}+{by}\right)\left({x}+{y}\right)={ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +\left({a}+{b}\right){xy} \\ $$$$\Rightarrow\mathrm{5}\left({x}+{y}\right)=\mathrm{10}+\left({a}+{b}\right){xy}\:\:\left({i}\right) \\ $$$$\left({ax}^{\mathrm{2}} +{by}^{\mathrm{2}} \right)\left({x}+{y}\right)={ax}^{\mathrm{3}} +{by}^{\mathrm{3}} +\left({ax}+{by}\right){xy} \\ $$$$\mathrm{10}\left({x}+{y}\right)=\mathrm{50}+\mathrm{5}{xy} \\ $$$$\Rightarrow\mathrm{2}\left({x}+{y}\right)=\mathrm{10}+{xy}\:\:\left({ii}\right) \\ $$$$\left({ax}^{\mathrm{3}} +{by}^{\mathrm{3}} \right)\left({x}+{y}\right)={ax}^{\mathrm{4}} +{by}^{\mathrm{4}} +\left({ax}^{\mathrm{2}} +{by}^{\mathrm{2}} \right){xy} \\ $$$$\mathrm{50}\left({x}+{y}\right)=\mathrm{130}+\mathrm{10}{xy} \\ $$$$\Rightarrow\mathrm{5}\left({x}+{y}\right)=\mathrm{13}+{xy}\:\:\left({iii}\right) \\ $$$$ \\ $$$$\Rightarrow{x}+{y}=\mathrm{1} \\ $$$$\Rightarrow{xy}=−\mathrm{8} \\ $$$$\Rightarrow{a}+{b}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$$$ \\ $$$$\mathrm{13}\left({x}+{y}−{xy}\right)−\mathrm{120}\left({a}+{b}\right) \\ $$$$=\mathrm{13}\left(\mathrm{1}+\mathrm{8}\right)−\mathrm{120}×\frac{\mathrm{5}}{\mathrm{8}}=\mathrm{42}\:\checkmark \\ $$
Commented by mr W last updated on 21/Jun/22
$${say}\:{p}_{{n}} ={ax}^{{n}} +{by}^{{n}} \\ $$$$\left({ax}^{{n}−\mathrm{1}} +{by}^{{n}−\mathrm{1}} \right)\left({x}+{y}\right)={ax}^{{n}} +{by}^{{n}} +\left({ax}^{{n}−\mathrm{2}} +{by}^{{n}−\mathrm{2}} \right){xy} \\ $$$$\left({x}+{y}\right){p}_{{n}−\mathrm{1}} ={p}_{{n}} +{xyp}_{{n}−\mathrm{2}} \\ $$$${x}+{y}=\mathrm{1},\:{xy}=−\mathrm{8} \\ $$$${p}_{{n}} ={p}_{{n}−\mathrm{1}} +\mathrm{8}{p}_{{n}−\mathrm{2}} \\ $$$${r}^{\mathrm{2}} −{r}−\mathrm{8}=\mathrm{0} \\ $$$${r}=\frac{\mathrm{1}\pm\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$$${p}_{{n}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} +{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \\ $$$${p}_{\mathrm{1}} ={A}+{B}=\mathrm{5} \\ $$$${p}_{\mathrm{2}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{2}}\right)+{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}}\right)=\mathrm{10} \\ $$$$\Rightarrow{A}=\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{33}}+\mathrm{3}}{\:\sqrt{\mathrm{33}}}\right) \\ $$$$\Rightarrow{B}=\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{33}}−\mathrm{3}}{\:\sqrt{\mathrm{33}}}\right) \\ $$$${p}_{{n}} =\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{33}}+\mathrm{3}}{\:\sqrt{\mathrm{33}}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} +\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{33}}−\mathrm{3}}{\:\sqrt{\mathrm{33}}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \\ $$$${p}_{{n}} =\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{33}}}\left\{\left(\sqrt{\mathrm{33}}+\mathrm{3}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} +\left(\sqrt{\mathrm{33}}−\mathrm{3}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \right\} \\ $$$${this}\:{is}\:{integerfor}\:{any}\:{n}\geqslant\mathrm{1}. \\ $$$${examples}: \\ $$$${p}_{\mathrm{5}} ={ax}^{\mathrm{5}} +{by}^{\mathrm{5}} =\mathrm{530} \\ $$$${p}_{\mathrm{6}} ={ax}^{\mathrm{6}} +{by}^{\mathrm{6}} =\mathrm{1570} \\ $$$${p}_{\mathrm{10}} ={ax}^{\mathrm{10}} +{by}^{\mathrm{10}} =\mathrm{211810} \\ $$$$…… \\ $$
Commented by Mikenice last updated on 23/Jun/22
$${thanks}\:{sir} \\ $$
Commented by Tawa11 last updated on 25/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$