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Question Number 87533 by mr W last updated on 04/Apr/20

Commented by MJS last updated on 05/Apr/20

a=((49)/(76))±((457(√(87)))/(6612)) b=((49)/(76))∓((457(√(87)))/(6612)) x=−7±(√(87)) y=−7∓(√(87)) answer is 20

$${a}=\frac{\mathrm{49}}{\mathrm{76}}\pm\frac{\mathrm{457}\sqrt{\mathrm{87}}}{\mathrm{6612}} \\ $$$${b}=\frac{\mathrm{49}}{\mathrm{76}}\mp\frac{\mathrm{457}\sqrt{\mathrm{87}}}{\mathrm{6612}} \\ $$$${x}=−\mathrm{7}\pm\sqrt{\mathrm{87}} \\ $$$${y}=−\mathrm{7}\mp\sqrt{\mathrm{87}} \\ $$$$\mathrm{answer}\:\mathrm{is}\:\mathrm{20} \\ $$

Commented by mr W last updated on 05/Apr/20

thanks sir!

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

Answered by ajfour last updated on 05/Apr/20

let ax=s, by=t ⇒ s+t=3 ....(i) sx+ty=7 ....(ii) sx^2 +ty^2 =16 ...(iii) sx^3 +ty^3 =42 ...(iv) Q=sx^4 +ty^4 =? (ii)−y×(i) s(x−y)=7−3y .....(A) (iii)−y×(ii) sx(x−y)=16−7y ....(B) ⇒ x=((16−7y)/(7−3y)) .....(I) (iii)−y^2 ×(i) s(x^2 −y^2 )=16−3y^2 ....(C) (iv)−y^2 ×(ii) sx(x^2 −y^2 )=42−7y^2 ....(D) ⇒ x=((42−7y^2 )/(16−3y^2 )) .....(II) equating (I) & (II) ((16−7y)/(7−3y))=((42−7y^2 )/(16−3y^2 )) 256−48y^2 −112y+21y^3 = 294−49y^2 −126y+21y^3 ⇒ y^2 +14y−38=0 y_1 =−7−(√(87)) , y_2 =−7+(√(87)) x_1 =((16−7y_1 )/(7−3y_1 ))=((16+7((√(87))+7))/(7+3((√(87))+7))) = ((65+7(√(87)))/(28+3(√(87)))) = −7+(√(87)) lets proceed with just x_1 , y_1 . s+t=3 sx+ty=7 ⇒ sx+(3−s)y=7 ⇒ s =((7−3y)/(x−y))=((7+3(7+(√(87))))/(2(√(87)))) = ((28+3(√(87)))/(2(√(87)))) = ((261+28(√(87)))/(174)) t=3−s = ((261−28(√(87)))/(174)) Q=sx^4 +ty^4 = ((261+28(√(87)))/(174))×{(√(87))−7}^4 +((261−28(√(87)))/(174))×{−(√(87))−7}^4 =2×10^1 .

$${let}\:\:\boldsymbol{{ax}}=\boldsymbol{{s}},\:\:\boldsymbol{{by}}=\boldsymbol{{t}} \\ $$$$\Rightarrow\:\:{s}+{t}=\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:....\left({i}\right) \\ $$$$\:\:\:\:\:\:\:{sx}+{ty}=\mathrm{7}\:\:\:\:\:\:....\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:{sx}^{\mathrm{2}} +{ty}^{\mathrm{2}} =\mathrm{16}\:\:\:\:...\left({iii}\right) \\ $$$$\:\:\:\:\:\:\:{sx}^{\mathrm{3}} +{ty}^{\mathrm{3}} =\mathrm{42}\:\:\:\:...\left({iv}\right) \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{Q}}={sx}^{\mathrm{4}} +{ty}^{\mathrm{4}} =? \\ $$$$\:\:\left({ii}\right)−{y}×\left({i}\right) \\ $$$$\:\:\:{s}\left({x}−{y}\right)=\mathrm{7}−\mathrm{3}{y}\:\:\:\:\:\:.....\left({A}\right) \\ $$$$\:\:\left({iii}\right)−{y}×\left({ii}\right) \\ $$$$\:\:\:{sx}\left({x}−{y}\right)=\mathrm{16}−\mathrm{7}{y}\:\:\:\:\:\:....\left({B}\right) \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{x}}=\frac{\mathrm{16}−\mathrm{7}\boldsymbol{{y}}}{\mathrm{7}−\mathrm{3}\boldsymbol{{y}}}\:\:\:\:\:\:\:\:\:\:\:\:.....\left(\boldsymbol{{I}}\right) \\ $$$$\:\left({iii}\right)−{y}^{\mathrm{2}} ×\left({i}\right) \\ $$$$\:\:{s}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{16}−\mathrm{3}{y}^{\mathrm{2}} \:\:\:\:\:\:\:....\left({C}\right) \\ $$$$\:\:\left({iv}\right)−{y}^{\mathrm{2}} ×\left({ii}\right) \\ $$$$\:\:{sx}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{42}−\mathrm{7}{y}^{\mathrm{2}} \:\:\:\:\:\:....\left({D}\right) \\ $$$$\Rightarrow\:\:\:\boldsymbol{{x}}=\frac{\mathrm{42}−\mathrm{7}\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{16}−\mathrm{3}\boldsymbol{{y}}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:.....\left(\boldsymbol{{II}}\right) \\ $$$${equating}\:\:\:\left({I}\right)\:\&\:\left({II}\right) \\ $$$$\:\:\:\frac{\mathrm{16}−\mathrm{7}\boldsymbol{{y}}}{\mathrm{7}−\mathrm{3}\boldsymbol{{y}}}=\frac{\mathrm{42}−\mathrm{7}\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{16}−\mathrm{3}\boldsymbol{{y}}^{\mathrm{2}} } \\ $$$$\mathrm{256}−\mathrm{48}{y}^{\mathrm{2}} −\mathrm{112}{y}+\mathrm{21}{y}^{\mathrm{3}} \\ $$$$\:\:\:\:=\:\mathrm{294}−\mathrm{49}{y}^{\mathrm{2}} −\mathrm{126}{y}+\mathrm{21}{y}^{\mathrm{3}} \\ $$$$ \\ $$$$\Rightarrow\:\:\boldsymbol{{y}}^{\mathrm{2}} +\mathrm{14}\boldsymbol{{y}}−\mathrm{38}=\mathrm{0} \\ $$$$\:\:\boldsymbol{{y}}_{\mathrm{1}} =−\mathrm{7}−\sqrt{\mathrm{87}}\:,\:\:\boldsymbol{{y}}_{\mathrm{2}} =−\mathrm{7}+\sqrt{\mathrm{87}} \\ $$$$\:\:\boldsymbol{{x}}_{\mathrm{1}} =\frac{\mathrm{16}−\mathrm{7}\boldsymbol{{y}}_{\mathrm{1}} }{\mathrm{7}−\mathrm{3}\boldsymbol{{y}}_{\mathrm{1}} }=\frac{\mathrm{16}+\mathrm{7}\left(\sqrt{\mathrm{87}}+\mathrm{7}\right)}{\mathrm{7}+\mathrm{3}\left(\sqrt{\mathrm{87}}+\mathrm{7}\right)} \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{65}+\mathrm{7}\sqrt{\mathrm{87}}}{\mathrm{28}+\mathrm{3}\sqrt{\mathrm{87}}}\:=\:−\mathrm{7}+\sqrt{\mathrm{87}} \\ $$$${lets}\:\:{proceed}\:{with}\:\:{just}\:{x}_{\mathrm{1}} ,\:{y}_{\mathrm{1}} . \\ $$$$\:\:\boldsymbol{{s}}+\boldsymbol{{t}}=\mathrm{3} \\ $$$$\:\:\boldsymbol{{sx}}+\boldsymbol{{ty}}=\mathrm{7}\:\:\:\Rightarrow\:\:{sx}+\left(\mathrm{3}−{s}\right){y}=\mathrm{7} \\ $$$$\Rightarrow\:\:\boldsymbol{{s}}\:=\frac{\mathrm{7}−\mathrm{3}{y}}{{x}−{y}}=\frac{\mathrm{7}+\mathrm{3}\left(\mathrm{7}+\sqrt{\mathrm{87}}\right)}{\mathrm{2}\sqrt{\mathrm{87}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{28}+\mathrm{3}\sqrt{\mathrm{87}}}{\mathrm{2}\sqrt{\mathrm{87}}}\:=\:\frac{\mathrm{261}+\mathrm{28}\sqrt{\mathrm{87}}}{\mathrm{174}} \\ $$$$\:\:\:\boldsymbol{{t}}=\mathrm{3}−\boldsymbol{{s}}\:=\:\frac{\mathrm{261}−\mathrm{28}\sqrt{\mathrm{87}}}{\mathrm{174}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{Q}}={sx}^{\mathrm{4}} +{ty}^{\mathrm{4}} \\ $$$$\:\:\:=\:\frac{\mathrm{261}+\mathrm{28}\sqrt{\mathrm{87}}}{\mathrm{174}}×\left\{\sqrt{\mathrm{87}}−\mathrm{7}\right\}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:+\frac{\mathrm{261}−\mathrm{28}\sqrt{\mathrm{87}}}{\mathrm{174}}×\left\{−\sqrt{\mathrm{87}}−\mathrm{7}\right\}^{\mathrm{4}} \\ $$$$\:\:=\mathrm{2}×\mathrm{10}^{\mathrm{1}} \:. \\ $$

Commented by mr W last updated on 05/Apr/20

thanks sir!

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

Commented by ajfour last updated on 05/Apr/20

Have you devised some still shorter better way sir, if so, please share..

$$\:{Have}\:{you}\:{devised}\:{some} \\ $$$${still}\:{shorter}\:{better}\:{way}\:{sir}, \\ $$$${if}\:{so},\:{please}\:{share}.. \\ $$

Commented by mr W last updated on 05/Apr/20

i have tried the same way as santu sir, i have no other better way. i′m considering how to get ax^n +by^n =? for n≥5, see an attempt below.

$${i}\:{have}\:{tried}\:{the}\:{same}\:{way}\:{as}\:{santu}\:{sir}, \\ $$$${i}\:{have}\:{no}\:{other}\:{better}\:{way}.\: \\ $$$${i}'{m}\:{considering}\:{how}\:{to}\:{get}\:\boldsymbol{{ax}}^{\boldsymbol{{n}}} +\boldsymbol{{by}}^{\boldsymbol{{n}}} =? \\ $$$${for}\:{n}\geqslant\mathrm{5},\:{see}\:{an}\:{attempt}\:{below}. \\ $$

Answered by john santu last updated on 05/Apr/20

(i)(ax^4 + by^4 )(x+y) = 42(x+y) ax^5 +b^y +ax^4 y + bxy^4 = 42(x+y) ax^5 +by^5 +xy(ax^3 +by^3 ) = 42(x+y) (ii)(ax^2 +by^2 )(x+y) = 7(x+y) ax^3 +by^3 +xy(ax+by) 7(x+y) = 16+xy .3 16(x+y) = 42 +xy.7 x+y = −14 ∧ xy = −38 ⇒ax^5 +by^5 = 42(−14)+(38)(16) = −588 + 608 = 20

$$\left(\mathrm{i}\right)\left(\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{by}^{\mathrm{4}} \right)\left(\mathrm{x}+\mathrm{y}\right)\:=\:\mathrm{42}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{ax}^{\mathrm{5}} +\mathrm{b}^{\mathrm{y}} \:+\mathrm{ax}^{\mathrm{4}} \mathrm{y}\:+\:\mathrm{bxy}^{\mathrm{4}} \:=\:\mathrm{42}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{ax}^{\mathrm{5}} +\mathrm{by}^{\mathrm{5}} +\mathrm{xy}\left(\mathrm{ax}^{\mathrm{3}} +\mathrm{by}^{\mathrm{3}} \right)\:=\:\mathrm{42}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\left(\mathrm{ii}\right)\left(\mathrm{ax}^{\mathrm{2}} +\mathrm{by}^{\mathrm{2}} \right)\left(\mathrm{x}+\mathrm{y}\right)\:=\:\mathrm{7}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{ax}^{\mathrm{3}} +\mathrm{by}^{\mathrm{3}} +\mathrm{xy}\left(\mathrm{ax}+\mathrm{by}\right)\: \\ $$$$\mathrm{7}\left(\mathrm{x}+\mathrm{y}\right)\:=\:\mathrm{16}+\mathrm{xy}\:.\mathrm{3} \\ $$$$\mathrm{16}\left(\mathrm{x}+\mathrm{y}\right)\:=\:\mathrm{42}\:+\mathrm{xy}.\mathrm{7} \\ $$$$\mathrm{x}+\mathrm{y}\:=\:−\mathrm{14}\:\wedge\:\mathrm{xy}\:=\:−\mathrm{38} \\ $$$$\Rightarrow\mathrm{ax}^{\mathrm{5}} +\mathrm{by}^{\mathrm{5}} \:=\:\mathrm{42}\left(−\mathrm{14}\right)+\left(\mathrm{38}\right)\left(\mathrm{16}\right) \\ $$$$=\:−\mathrm{588}\:+\:\mathrm{608}\:=\:\mathrm{20}\: \\ $$

Commented by ajfour last updated on 05/Apr/20

Superb Sir! Congratulations, ingenious way.

$$\mathcal{S}{uperb}\:\mathcal{S}{ir}!\:{Congratulations}, \\ $$$${ingenious}\:{way}. \\ $$

Commented by jagoll last updated on 05/Apr/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 05/Apr/20

thanks sir!

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

Commented by mr W last updated on 05/Apr/20

i think this is the best way. but can we apply it to get ax^n +by^n generally? i think this is possible, see below.

$${i}\:{think}\:{this}\:{is}\:{the}\:{best}\:{way}.\:{but}\:{can}\:{we} \\ $$$${apply}\:{it}\:{to}\:{get}\:\boldsymbol{{ax}}^{\boldsymbol{{n}}} +\boldsymbol{{by}}^{\boldsymbol{{n}}} \:{generally}? \\ $$$${i}\:{think}\:{this}\:{is}\:{possible},\:{see}\:{below}. \\ $$

Answered by mr W last updated on 26/Oct/21

S_n =ax^n +by^n S_1 =3 S_2 =7 S_3 =16 S_4 =42 S_(n+1) =ax^(n+1) +by^(n+1) (x+y)S_n =(x+y)(ax^n +by^n )=ax^(n+1) +by^(n+1) +xy(ax^(n−1) +by^(n−1) ) (x+y)S_n =S_(n+1) +xyS_(n−1) ⇒S_(n+1) =(x+y)S_n −xyS_(n−1) let x+y=p, xy=q ⇒S_(n+1) =pS_n −qS_(n−1) ...(i) S_3 =pS_2 −qS_1 S_4 =pS_3 −qS_2 ⇒p=((S_1 S_4 −S_2 S_3 )/(S_1 S_3 −S_2 ^2 ))=((3×42−7×16)/(3×16−7^2 ))=−14 ⇒q=((S_2 S_4 −S_3 ^2 )/(S_1 S_3 −S_2 ^2 ))=((7×42−16^2 )/(3×16−7^2 ))=−38 a=((S_2 −S_1 y)/(x(x−y))) b=((S_2 −S_1 x)/(y(y−x))) S_2 =pS_1 −qS_0 ⇒S_0 =a+b=((pS_1 −S_2 )/q) from (i) we have the characteristic eqn. t^2 −pt+q=0 t=((p±(√(p^2 −4q)))/2) let α=((p+(√(p^2 −4q)))/2), β=((p−(√(p^2 −4q)))/2) ⇒S_n =Aα^n +Bβ^n ...(ii) S_1 =Aα+Bβ ...(iii) S_2 =Aα^2 +Bβ^2 ...(iv) ⇒A=((βS_1 −S_2 )/(α(β−α))) ⇒B=((αS_1 −S_2 )/(β(α−β))) with our example: α=((−14+(√(14^2 +4×38)))/2)=−7+(√(87)) β=((−14−(√(14^2 +4×38)))/2)=−7−(√(87)) A=(1/(76))(49+((457(√(87)))/(87))) B=(1/(76))(49−((457(√(87)))/(87))) S_n =(1/(76)){(49+((457(√(87)))/(87)))(−7+(√(87)))^n +(49+((457(√(87)))/(87)))(−7−(√(87)))^n } we get: S_5 =20 S_6 =1316 S_7 =−17446 S_8 =297304 S_9 =−4833488 S_(10) =79966384

$${S}_{{n}} ={ax}^{{n}} +{by}^{{n}} \\ $$$${S}_{\mathrm{1}} =\mathrm{3} \\ $$$${S}_{\mathrm{2}} =\mathrm{7} \\ $$$${S}_{\mathrm{3}} =\mathrm{16} \\ $$$${S}_{\mathrm{4}} =\mathrm{42} \\ $$$${S}_{{n}+\mathrm{1}} ={ax}^{{n}+\mathrm{1}} +{by}^{{n}+\mathrm{1}} \\ $$$$\left({x}+{y}\right){S}_{{n}} =\left({x}+{y}\right)\left({ax}^{{n}} +{by}^{{n}} \right)={ax}^{{n}+\mathrm{1}} +{by}^{{n}+\mathrm{1}} +{xy}\left({ax}^{{n}−\mathrm{1}} +{by}^{{n}−\mathrm{1}} \right) \\ $$$$\left({x}+{y}\right){S}_{{n}} ={S}_{{n}+\mathrm{1}} +{xyS}_{{n}−\mathrm{1}} \\ $$$$\Rightarrow{S}_{{n}+\mathrm{1}} =\left({x}+{y}\right){S}_{{n}} −{xyS}_{{n}−\mathrm{1}} \\ $$$${let}\:{x}+{y}={p},\:{xy}={q} \\ $$$$\Rightarrow{S}_{{n}+\mathrm{1}} ={pS}_{{n}} −{qS}_{{n}−\mathrm{1}} \:\:\:\:\:...\left({i}\right) \\ $$$${S}_{\mathrm{3}} ={pS}_{\mathrm{2}} −{qS}_{\mathrm{1}} \\ $$$${S}_{\mathrm{4}} ={pS}_{\mathrm{3}} −{qS}_{\mathrm{2}} \\ $$$$\Rightarrow{p}=\frac{{S}_{\mathrm{1}} {S}_{\mathrm{4}} −{S}_{\mathrm{2}} {S}_{\mathrm{3}} }{{S}_{\mathrm{1}} {S}_{\mathrm{3}} −{S}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{\mathrm{3}×\mathrm{42}−\mathrm{7}×\mathrm{16}}{\mathrm{3}×\mathrm{16}−\mathrm{7}^{\mathrm{2}} }=−\mathrm{14} \\ $$$$\Rightarrow{q}=\frac{{S}_{\mathrm{2}} {S}_{\mathrm{4}} −{S}_{\mathrm{3}} ^{\mathrm{2}} }{{S}_{\mathrm{1}} {S}_{\mathrm{3}} −{S}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{\mathrm{7}×\mathrm{42}−\mathrm{16}^{\mathrm{2}} }{\mathrm{3}×\mathrm{16}−\mathrm{7}^{\mathrm{2}} }=−\mathrm{38} \\ $$$$ \\ $$$${a}=\frac{{S}_{\mathrm{2}} −{S}_{\mathrm{1}} {y}}{{x}\left({x}−{y}\right)} \\ $$$${b}=\frac{{S}_{\mathrm{2}} −{S}_{\mathrm{1}} {x}}{{y}\left({y}−{x}\right)} \\ $$$$ \\ $$$${S}_{\mathrm{2}} ={pS}_{\mathrm{1}} −{qS}_{\mathrm{0}} \\ $$$$\Rightarrow{S}_{\mathrm{0}} ={a}+{b}=\frac{{pS}_{\mathrm{1}} −{S}_{\mathrm{2}} }{{q}} \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{we}\:{have}\:{the}\:{characteristic}\:{eqn}. \\ $$$${t}^{\mathrm{2}} −{pt}+{q}=\mathrm{0} \\ $$$${t}=\frac{{p}\pm\sqrt{{p}^{\mathrm{2}} −\mathrm{4}{q}}}{\mathrm{2}} \\ $$$${let}\:\alpha=\frac{{p}+\sqrt{{p}^{\mathrm{2}} −\mathrm{4}{q}}}{\mathrm{2}},\:\beta=\frac{{p}−\sqrt{{p}^{\mathrm{2}} −\mathrm{4}{q}}}{\mathrm{2}} \\ $$$$\Rightarrow{S}_{{n}} ={A}\alpha^{{n}} +{B}\beta^{{n}} \:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${S}_{\mathrm{1}} ={A}\alpha+{B}\beta\:\:\:...\left({iii}\right) \\ $$$${S}_{\mathrm{2}} ={A}\alpha^{\mathrm{2}} +{B}\beta^{\mathrm{2}} \:\:\:...\left({iv}\right) \\ $$$$\Rightarrow{A}=\frac{\beta{S}_{\mathrm{1}} −{S}_{\mathrm{2}} }{\alpha\left(\beta−\alpha\right)} \\ $$$$\Rightarrow{B}=\frac{\alpha{S}_{\mathrm{1}} −{S}_{\mathrm{2}} }{\beta\left(\alpha−\beta\right)} \\ $$$$ \\ $$$${with}\:{our}\:{example}: \\ $$$$\alpha=\frac{−\mathrm{14}+\sqrt{\mathrm{14}^{\mathrm{2}} +\mathrm{4}×\mathrm{38}}}{\mathrm{2}}=−\mathrm{7}+\sqrt{\mathrm{87}} \\ $$$$\beta=\frac{−\mathrm{14}−\sqrt{\mathrm{14}^{\mathrm{2}} +\mathrm{4}×\mathrm{38}}}{\mathrm{2}}=−\mathrm{7}−\sqrt{\mathrm{87}} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{76}}\left(\mathrm{49}+\frac{\mathrm{457}\sqrt{\mathrm{87}}}{\mathrm{87}}\right) \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{76}}\left(\mathrm{49}−\frac{\mathrm{457}\sqrt{\mathrm{87}}}{\mathrm{87}}\right) \\ $$$$ \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{76}}\left\{\left(\mathrm{49}+\frac{\mathrm{457}\sqrt{\mathrm{87}}}{\mathrm{87}}\right)\left(−\mathrm{7}+\sqrt{\mathrm{87}}\right)^{{n}} +\left(\mathrm{49}+\frac{\mathrm{457}\sqrt{\mathrm{87}}}{\mathrm{87}}\right)\left(−\mathrm{7}−\sqrt{\mathrm{87}}\right)^{{n}} \right\} \\ $$$${we}\:{get}: \\ $$$${S}_{\mathrm{5}} =\mathrm{20} \\ $$$${S}_{\mathrm{6}} =\mathrm{1316} \\ $$$${S}_{\mathrm{7}} =−\mathrm{17446} \\ $$$${S}_{\mathrm{8}} =\mathrm{297304} \\ $$$${S}_{\mathrm{9}} =−\mathrm{4833488} \\ $$$${S}_{\mathrm{10}} =\mathrm{79966384} \\ $$

Commented by john santu last updated on 05/Apr/20

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 05/Apr/20

recursively we can get S_5 =−14S_4 +38S_3 =−14×42+38×16=20 S_6 =−14S_5 +38S_4 =−14×20+38×42=1316 S_7 =−14S_6 +38S_5 =−14×1316+38×20=−17664 ...

$${recursively}\:{we}\:{can}\:{get} \\ $$$${S}_{\mathrm{5}} =−\mathrm{14}{S}_{\mathrm{4}} +\mathrm{38}{S}_{\mathrm{3}} =−\mathrm{14}×\mathrm{42}+\mathrm{38}×\mathrm{16}=\mathrm{20} \\ $$$${S}_{\mathrm{6}} =−\mathrm{14}{S}_{\mathrm{5}} +\mathrm{38}{S}_{\mathrm{4}} =−\mathrm{14}×\mathrm{20}+\mathrm{38}×\mathrm{42}=\mathrm{1316} \\ $$$${S}_{\mathrm{7}} =−\mathrm{14}{S}_{\mathrm{6}} +\mathrm{38}{S}_{\mathrm{5}} =−\mathrm{14}×\mathrm{1316}+\mathrm{38}×\mathrm{20}=−\mathrm{17664} \\ $$$$... \\ $$

Commented by TawaTawa1 last updated on 05/Apr/20

Wow great sir. God bless you sir.

$$\mathrm{Wow}\:\mathrm{great}\:\mathrm{sir}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by TawaTawa1 last updated on 05/Apr/20

Sir, what of if x ± (1/x) = a Then x^n ± (1/x^n ) = ??? What is the general form sir?

$$\mathrm{Sir},\:\:\mathrm{what}\:\mathrm{of}\:\:\:\mathrm{if}\:\:\:\:\:\:\:\:\mathrm{x}\:\:\pm\:\:\frac{\mathrm{1}}{\mathrm{x}}\:\:\:=\:\:\mathrm{a} \\ $$$$\mathrm{Then}\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{n}} \:\:\pm\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{n}} }\:\:\:=\:\:\:??? \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{general}\:\mathrm{form}\:\mathrm{sir}? \\ $$

Commented by mr W last updated on 05/Apr/20

this question is already answered n times, please try to find an old post concerning this question. the solution is also easy: get x from x±(1/x)=a, this should be solvable. then you can get x^n ±(1/x^n ).

$${this}\:{question}\:{is}\:{already}\:{answered}\:{n} \\ $$$${times},\:{please}\:{try}\:{to}\:{find}\:{an}\:{old}\:{post} \\ $$$${concerning}\:{this}\:{question}.\:{the}\:{solution} \\ $$$${is}\:{also}\:{easy}:\:{get}\:{x}\:{from}\:{x}\pm\frac{\mathrm{1}}{{x}}={a},\:{this} \\ $$$${should}\:{be}\:{solvable}.\:{then}\:{you}\:{can}\:{get} \\ $$$${x}^{{n}} \pm\frac{\mathrm{1}}{{x}^{{n}} }. \\ $$

Commented by TawaTawa1 last updated on 05/Apr/20

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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