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Question Number 171861 by Mikenice last updated on 21/Jun/22

if ax+by=5 ax^2 +by^2 =10 ax^3 +by^3 =50 ax^4 +by^4 =130 find 13(x+y−xy)−120(a+b)

$${if}\: \\ $$$${ax}+{by}=\mathrm{5} \\ $$$${ax}^{\mathrm{2}} +{by}^{\mathrm{2}} =\mathrm{10} \\ $$$${ax}^{\mathrm{3}} +{by}^{\mathrm{3}} =\mathrm{50} \\ $$$${ax}^{\mathrm{4}} +{by}^{\mathrm{4}} =\mathrm{130} \\ $$$${find}\:\mathrm{13}\left({x}+{y}−{xy}\right)−\mathrm{120}\left({a}+{b}\right) \\ $$

Commented by infinityaction last updated on 21/Jun/22

42

$$\mathrm{42} \\ $$

Commented by Mikenice last updated on 21/Jun/22

please sir show the solution

$${please}\:{sir}\:{show}\:{the}\:{solution} \\ $$

Commented by mr W last updated on 21/Jun/22

it′s more interesting and challenging to find ax^5 +by^5 , ax^6 +by^6 or generally ax^n +by^n =?

$${it}'{s}\:{more}\:{interesting}\:{and} \\ $$$${challenging}\:{to}\:{find}\:{ax}^{\mathrm{5}} +{by}^{\mathrm{5}} ,\:{ax}^{\mathrm{6}} +{by}^{\mathrm{6}} \\ $$$${or}\:{generally}\:{ax}^{{n}} +{by}^{{n}} =? \\ $$

Commented by Mikenice last updated on 21/Jun/22

if you can show solution, please do so

$${if}\:{you}\:{can}\:{show}\:{solution},\:{please}\:{do}\:{so} \\ $$

Commented by mr W last updated on 21/Jun/22

(ax+by)(x+y)=ax^2 +by^2 +(a+b)xy ⇒5(x+y)=10+(a+b)xy (i) (ax^2 +by^2 )(x+y)=ax^3 +by^3 +(ax+by)xy 10(x+y)=50+5xy ⇒2(x+y)=10+xy (ii) (ax^3 +by^3 )(x+y)=ax^4 +by^4 +(ax^2 +by^2 )xy 50(x+y)=130+10xy ⇒5(x+y)=13+xy (iii) ⇒x+y=1 ⇒xy=−8 ⇒a+b=(5/8) 13(x+y−xy)−120(a+b) =13(1+8)−120×(5/8)=42 ✓

$$\left({ax}+{by}\right)\left({x}+{y}\right)={ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +\left({a}+{b}\right){xy} \\ $$$$\Rightarrow\mathrm{5}\left({x}+{y}\right)=\mathrm{10}+\left({a}+{b}\right){xy}\:\:\left({i}\right) \\ $$$$\left({ax}^{\mathrm{2}} +{by}^{\mathrm{2}} \right)\left({x}+{y}\right)={ax}^{\mathrm{3}} +{by}^{\mathrm{3}} +\left({ax}+{by}\right){xy} \\ $$$$\mathrm{10}\left({x}+{y}\right)=\mathrm{50}+\mathrm{5}{xy} \\ $$$$\Rightarrow\mathrm{2}\left({x}+{y}\right)=\mathrm{10}+{xy}\:\:\left({ii}\right) \\ $$$$\left({ax}^{\mathrm{3}} +{by}^{\mathrm{3}} \right)\left({x}+{y}\right)={ax}^{\mathrm{4}} +{by}^{\mathrm{4}} +\left({ax}^{\mathrm{2}} +{by}^{\mathrm{2}} \right){xy} \\ $$$$\mathrm{50}\left({x}+{y}\right)=\mathrm{130}+\mathrm{10}{xy} \\ $$$$\Rightarrow\mathrm{5}\left({x}+{y}\right)=\mathrm{13}+{xy}\:\:\left({iii}\right) \\ $$$$ \\ $$$$\Rightarrow{x}+{y}=\mathrm{1} \\ $$$$\Rightarrow{xy}=−\mathrm{8} \\ $$$$\Rightarrow{a}+{b}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$$$ \\ $$$$\mathrm{13}\left({x}+{y}−{xy}\right)−\mathrm{120}\left({a}+{b}\right) \\ $$$$=\mathrm{13}\left(\mathrm{1}+\mathrm{8}\right)−\mathrm{120}×\frac{\mathrm{5}}{\mathrm{8}}=\mathrm{42}\:\checkmark \\ $$

Commented by mr W last updated on 21/Jun/22

say p_n =ax^n +by^n (ax^(n−1) +by^(n−1) )(x+y)=ax^n +by^n +(ax^(n−2) +by^(n−2) )xy (x+y)p_(n−1) =p_n +xyp_(n−2) x+y=1, xy=−8 p_n =p_(n−1) +8p_(n−2) r^2 −r−8=0 r=((1±(√(33)))/2) p_n =A(((1+(√(33)))/2))^(n−1) +B(((1−(√(33)))/2))^(n−1) p_1 =A+B=5 p_2 =A(((1+(√(33)))/2))+B(((1−(√(33)))/2))=10 ⇒A=(5/2)((((√(33))+3)/( (√(33))))) ⇒B=(5/2)((((√(33))−3)/( (√(33))))) p_n =(5/2)((((√(33))+3)/( (√(33)))))(((1+(√(33)))/2))^(n−1) +(5/2)((((√(33))−3)/( (√(33)))))(((1−(√(33)))/2))^(n−1) p_n =(5/(2(√(33)))){((√(33))+3)(((1+(√(33)))/2))^(n−1) +((√(33))−3)(((1−(√(33)))/2))^(n−1) } this is integerfor any n≥1. examples: p_5 =ax^5 +by^5 =530 p_6 =ax^6 +by^6 =1570 p_(10) =ax^(10) +by^(10) =211810 ......

$${say}\:{p}_{{n}} ={ax}^{{n}} +{by}^{{n}} \\ $$$$\left({ax}^{{n}−\mathrm{1}} +{by}^{{n}−\mathrm{1}} \right)\left({x}+{y}\right)={ax}^{{n}} +{by}^{{n}} +\left({ax}^{{n}−\mathrm{2}} +{by}^{{n}−\mathrm{2}} \right){xy} \\ $$$$\left({x}+{y}\right){p}_{{n}−\mathrm{1}} ={p}_{{n}} +{xyp}_{{n}−\mathrm{2}} \\ $$$${x}+{y}=\mathrm{1},\:{xy}=−\mathrm{8} \\ $$$${p}_{{n}} ={p}_{{n}−\mathrm{1}} +\mathrm{8}{p}_{{n}−\mathrm{2}} \\ $$$${r}^{\mathrm{2}} −{r}−\mathrm{8}=\mathrm{0} \\ $$$${r}=\frac{\mathrm{1}\pm\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$$${p}_{{n}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} +{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \\ $$$${p}_{\mathrm{1}} ={A}+{B}=\mathrm{5} \\ $$$${p}_{\mathrm{2}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{2}}\right)+{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}}\right)=\mathrm{10} \\ $$$$\Rightarrow{A}=\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{33}}+\mathrm{3}}{\:\sqrt{\mathrm{33}}}\right) \\ $$$$\Rightarrow{B}=\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{33}}−\mathrm{3}}{\:\sqrt{\mathrm{33}}}\right) \\ $$$${p}_{{n}} =\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{33}}+\mathrm{3}}{\:\sqrt{\mathrm{33}}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} +\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{33}}−\mathrm{3}}{\:\sqrt{\mathrm{33}}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \\ $$$${p}_{{n}} =\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{33}}}\left\{\left(\sqrt{\mathrm{33}}+\mathrm{3}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} +\left(\sqrt{\mathrm{33}}−\mathrm{3}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \right\} \\ $$$${this}\:{is}\:{integerfor}\:{any}\:{n}\geqslant\mathrm{1}. \\ $$$${examples}: \\ $$$${p}_{\mathrm{5}} ={ax}^{\mathrm{5}} +{by}^{\mathrm{5}} =\mathrm{530} \\ $$$${p}_{\mathrm{6}} ={ax}^{\mathrm{6}} +{by}^{\mathrm{6}} =\mathrm{1570} \\ $$$${p}_{\mathrm{10}} ={ax}^{\mathrm{10}} +{by}^{\mathrm{10}} =\mathrm{211810} \\ $$$$...... \\ $$

Commented by Mikenice last updated on 23/Jun/22

thanks sir

$${thanks}\:{sir} \\ $$

Commented by Tawa11 last updated on 25/Jun/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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