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Question Number 195854 by sonukgindia last updated on 11/Aug/23
Commented by Rasheed.Sindhi last updated on 13/Aug/23
Q#194637
$${Q}#\mathrm{194637} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Aug/23
Q#191527
$${Q}#\mathrm{191527} \\ $$
Answered by MM42 last updated on 12/Aug/23
p^2 +q^2 =(p+q)^2 −2pq⇒pq=−(1/2) p^3 +q^3 =(p+q)^3 −3pq(p+q)=(5/2) p^4 +q^4 =(p^2 +q^2 )^2 −2p^2 q^2 =(7/2) p^5 +q^5 =(p+q)(p^4 +q^4 )−pq(p^3 +q^3 )=((19)/4) p^6 +q^6 =(p^3 +q^3 )^2 −2p^3 q^3 =((25)/4)+(1/4)=((13)/2) p^(11) +q^(11) =(p^6 +q^6 )(p^5 +q^5 )−p^5 q^5 (p+q)=((989)/(32)) ✓
$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq}\Rightarrow{pq}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${p}^{\mathrm{3}} +{q}^{\mathrm{3}} =\left({p}+{q}\right)^{\mathrm{3}} −\mathrm{3}{pq}\left({p}+{q}\right)=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${p}^{\mathrm{4}} +{q}^{\mathrm{4}} =\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${p}^{\mathrm{5}} +{q}^{\mathrm{5}} =\left({p}+{q}\right)\left({p}^{\mathrm{4}} +{q}^{\mathrm{4}} \right)−{pq}\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)=\frac{\mathrm{19}}{\mathrm{4}} \\ $$$${p}^{\mathrm{6}} +{q}^{\mathrm{6}} =\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}{p}^{\mathrm{3}} {q}^{\mathrm{3}} =\frac{\mathrm{25}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{13}}{\mathrm{2}} \\ $$$${p}^{\mathrm{11}} +{q}^{\mathrm{11}} =\left({p}^{\mathrm{6}} +{q}^{\mathrm{6}} \right)\left({p}^{\mathrm{5}} +{q}^{\mathrm{5}} \right)−{p}^{\mathrm{5}} {q}^{\mathrm{5}} \left({p}+{q}\right)=\frac{\mathrm{989}}{\mathrm{32}}\:\checkmark \\ $$$$ \\ $$
Answered by mr W last updated on 12/Aug/23
METHOD I (p+q)^2 =p^2 +q^2 +2pq 1^2 =2+2pq ⇒pq=−(1/2) (p^2 +q^2 )(p+q)=p^3 +q^3 +pq(p+q) 2×1=p^3 +q^3 −(1/2)×1 ⇒p^3 +q^3 =(5/2) (p^3 +q^3 )^3 =p^9 +q^9 +3(pq)^3 (p^3 +q^3 ) ((5/2))^3 =p^9 +q^9 +3(−(1/2))^3 ((5/2)) ⇒p^9 +q^9 =((265)/(16)) (p^2 +q^2 )^2 =p^4 +q^4 +2(pq)^2 2^2 =p^4 +q^4 +2(−(1/2))^2 ⇒p^4 +q^4 =(7/2) (p^4 +q^4 )(p+q)=p^5 +q^5 +pq(p^3 +q^3 ) (7/2)×1=p^5 +q^5 −(1/2)×(5/2) ⇒p^5 +q^5 =((19)/4) (p^5 +q^5 )^2 =p^(10) +q^(10) +2(pq)^5 (((19)/4))^2 =p^(10) +q^(10) +2(−(1/2))^5 ⇒p^(10) +q^(10) =((181)/8) (p^(10) +q^(10) )(p+q)=p^(11) +q^(11) +pq(p^9 +q^9 ) (((181)/8))×1=p^(11) +q^(11) −(1/2)×((265)/(16)) ⇒p^(11) +q^(11) =((989)/(32)) ✓
$$\underline{{METHOD}\:{I}} \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\mathrm{2}{pq} \\ $$$$\mathrm{1}^{\mathrm{2}} =\mathrm{2}+\mathrm{2}{pq}\: \\ $$$$\Rightarrow{pq}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left({p}+{q}\right)={p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{pq}\left({p}+{q}\right) \\ $$$$\mathrm{2}×\mathrm{1}={p}^{\mathrm{3}} +{q}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1} \\ $$$$\Rightarrow{p}^{\mathrm{3}} +{q}^{\mathrm{3}} =\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)^{\mathrm{3}} ={p}^{\mathrm{9}} +{q}^{\mathrm{9}} +\mathrm{3}\left({pq}\right)^{\mathrm{3}} \left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right) \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{3}} ={p}^{\mathrm{9}} +{q}^{\mathrm{9}} +\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \left(\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{p}^{\mathrm{9}} +{q}^{\mathrm{9}} =\frac{\mathrm{265}}{\mathrm{16}} \\ $$$$\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{\mathrm{2}} ={p}^{\mathrm{4}} +{q}^{\mathrm{4}} +\mathrm{2}\left({pq}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{2}} ={p}^{\mathrm{4}} +{q}^{\mathrm{4}} +\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{p}^{\mathrm{4}} +{q}^{\mathrm{4}} =\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{4}} +{q}^{\mathrm{4}} \right)\left({p}+{q}\right)={p}^{\mathrm{5}} +{q}^{\mathrm{5}} +{pq}\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right) \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}×\mathrm{1}={p}^{\mathrm{5}} +{q}^{\mathrm{5}} −\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow{p}^{\mathrm{5}} +{q}^{\mathrm{5}} =\frac{\mathrm{19}}{\mathrm{4}} \\ $$$$\left({p}^{\mathrm{5}} +{q}^{\mathrm{5}} \right)^{\mathrm{2}} ={p}^{\mathrm{10}} +{q}^{\mathrm{10}} +\mathrm{2}\left({pq}\right)^{\mathrm{5}} \\ $$$$\left(\frac{\mathrm{19}}{\mathrm{4}}\right)^{\mathrm{2}} ={p}^{\mathrm{10}} +{q}^{\mathrm{10}} +\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{5}} \\ $$$$\Rightarrow{p}^{\mathrm{10}} +{q}^{\mathrm{10}} =\frac{\mathrm{181}}{\mathrm{8}} \\ $$$$\left({p}^{\mathrm{10}} +{q}^{\mathrm{10}} \right)\left({p}+{q}\right)={p}^{\mathrm{11}} +{q}^{\mathrm{11}} +{pq}\left({p}^{\mathrm{9}} +{q}^{\mathrm{9}} \right) \\ $$$$\left(\frac{\mathrm{181}}{\mathrm{8}}\right)×\mathrm{1}={p}^{\mathrm{11}} +{q}^{\mathrm{11}} −\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{265}}{\mathrm{16}} \\ $$$$\Rightarrow{p}^{\mathrm{11}} +{q}^{\mathrm{11}} =\frac{\mathrm{989}}{\mathrm{32}}\:\checkmark \\ $$
Answered by mr W last updated on 12/Aug/23
METHOD II let s_n =p^n +q^n s_1 =p+q=e_1 =1 s_2 =p^2 +q^2 =e_1 p_1 −2e_2 2=1×1−2e_2 ⇒e_2 =−(1/2) s_n =e_1 s_(n−1) −e_2 s_(n−2) s_n −s_(n−1) −(1/2)s_(n−2) =0 r^2 −r−(1/2)=0 ⇒r_(1,2) =((1±(√3))/2) ⇒s_n =(((1+(√3))/2))^n +(((1−(√3))/2))^n =((𝚺_(k=0) ^(⌊(n/2)⌋) 3^k C_(2k) ^n )/2^(n−1) ) s_(11) =((989)/(32)) ✓ more examples: s_(20) =((268377088)/2^(19) )=((262 087)/(512)) s_(40) =((144052522725670912)/2^(39) )=((137 379 291 137)/(524 288))
$$\underline{{METHOD}\:{II}} \\ $$$${let}\:\boldsymbol{{s}}_{\boldsymbol{{n}}} =\boldsymbol{{p}}^{\boldsymbol{{n}}} +\boldsymbol{{q}}^{\boldsymbol{{n}}} \\ $$$${s}_{\mathrm{1}} ={p}+{q}={e}_{\mathrm{1}} =\mathrm{1} \\ $$$${s}_{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} \\ $$$$\mathrm{2}=\mathrm{1}×\mathrm{1}−\mathrm{2}{e}_{\mathrm{2}} \:\Rightarrow{e}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${s}_{{n}} ={e}_{\mathrm{1}} {s}_{{n}−\mathrm{1}} −{e}_{\mathrm{2}} {s}_{{n}−\mathrm{2}} \\ $$$$\boldsymbol{{s}}_{\boldsymbol{{n}}} −\boldsymbol{{s}}_{\boldsymbol{{n}}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{s}}_{\boldsymbol{{n}}−\mathrm{2}} =\mathrm{0} \\ $$$${r}^{\mathrm{2}} −{r}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{r}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{{s}}_{\boldsymbol{{n}}} =\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\boldsymbol{{n}}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\boldsymbol{{n}}} =\frac{\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\lfloor\frac{\boldsymbol{{n}}}{\mathrm{2}}\rfloor} {\boldsymbol{\sum}}}\mathrm{3}^{\boldsymbol{{k}}} \boldsymbol{{C}}_{\mathrm{2}\boldsymbol{{k}}} ^{\boldsymbol{{n}}} }{\mathrm{2}^{\boldsymbol{{n}}−\mathrm{1}} } \\ $$$${s}_{\mathrm{11}} =\frac{\mathrm{989}}{\mathrm{32}}\:\checkmark \\ $$$${more}\:{examples}: \\ $$$${s}_{\mathrm{20}} =\frac{\mathrm{268377088}}{\mathrm{2}^{\mathrm{19}} }=\frac{\mathrm{262}\:\mathrm{087}}{\mathrm{512}} \\ $$$${s}_{\mathrm{40}} =\frac{\mathrm{144052522725670912}}{\mathrm{2}^{\mathrm{39}} }=\frac{\mathrm{137}\:\mathrm{379}\:\mathrm{291}\:\mathrm{137}}{\mathrm{524}\:\mathrm{288}} \\ $$
Commented by jabarsing last updated on 12/Aug/23
verh very nice
$${verh}\:{very}\:{nice} \\ $$
Commented by MM42 last updated on 12/Aug/23
it was great
$${it}\:{was}\:{great} \\ $$

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