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Question Number 81308 by naka3546 last updated on 11/Feb/20

Commented by mr W last updated on 11/Feb/20

2104?

2104?

Commented by naka3546 last updated on 11/Feb/20

please, show  your  working, sir.

please,showyourworking,sir.

Answered by mr W last updated on 12/Feb/20

given is the recursive relation  x_(n+13) =x_(n+4) +2x_n   we see x_n =Aq^n  fulfills this condition.  Aq^(n+13) =Aq^(n+4) +2Aq^n   ⇒q^(13) =q^4 +2  it has 13 roots, say q_r  (r=1,2,...,13).  the general formula for term x_n  is then  x_n =Σ_(r=1) ^(13) A_r q_r ^n   with q_r ^(13) =q_r ^4 +2   (r=1,2,3,...,13)    x_0 =Σ_(r=1) ^(13) A_r =1  (x_(13) =x_4 +2x_0  ⇒x_0 =1)  x_m =Σ_(r=1) ^(13) A_r q_r ^m =0   (m=1,2,3,...,12)  x_(13) =Σ_(r=1) ^(13) A_r q_r ^(13) =2    x_(143) =Σ_(r=1) ^(13) A_r q_r ^(143)   q_r ^(143) =(q_r ^(13) )^(11) =(q_r ^4 +2)^(11) =Σ_(k=0) ^(11) C_k ^(11) q_r ^(4k) 2^(11−k)     x_(143) =Σ_(r=1) ^(13) A_r (Σ_(k=0) ^(11) C_k ^(11) q_r ^(4k) 2^(11−k) )  x_(143) =Σ_(k=0) ^(11) C_k ^(11) 2^(11−k) (Σ_(r=1) ^(13) A_r q_r ^(4k) )  k=0 ⇒ Σ_(r=1) ^(13) A_r q_r ^(4k) =1  k=1..3 ⇒ Σ_(r=1) ^(13) A_r q_r ^(4k) =0  ⇒x_(134) =2^(11) +Σ_(k=4) ^(11) C_k ^(11) 2^(11−k) (Σ_(r=1) ^(13) A_r q_r ^(4k) )  k=4..11:  q_r ^(16) =q_r ^(13) q_r ^3 =(q_r ^4 +2)q_r ^3 =q_r ^7 +2q_r ^3   ⇒Σ_(r=1) ^(13) A_r q_r ^(16) =Σ_(r=1) ^(13) A_r q_r ^7 +2Σ_(r=1) ^(13) A_r q_r ^3 =0  q_r ^(20) =q_r ^(13) q_r ^7 =(q_r ^4 +2)q_r ^7 =q_r ^(11) +2q_r ^7   ⇒Σ_(r=1) ^(13) A_r q_r ^(20) =Σ_(r=1) ^(13) A_r q_r ^(11) +2Σ_(r=1) ^(13) A_r q_r ^7 =0  q_r ^(24) =q_r ^(13) q_r ^(11) =(q_r ^4 +2)q_r ^(11) =(q_r ^4 +2)q_r ^2 +2q_r ^(11)   ⇒Σ_(r=1) ^(13) A_r q_r ^(24) =0  q_r ^(28) =(q_r ^(13) )^2 q_r ^2 =(q_r ^4 +2)^2 q_r ^2 =q_r ^(10) +4q_r ^6 +4q_r ^2   ⇒Σ_(r=1) ^(13) A_r q_r ^(28) =0  q_r ^(32) =(q_r ^(13) )^2 q_r ^6 =(q_r ^4 +2)^2 q_r ^6 =q_r ^(14) +4q_r ^(10) +4q_r ^6   =(q_r ^4 +2)q_r +4q_r ^(10) +4q_r ^6   ⇒Σ_(r=1) ^(13) A_r q_r ^(32) =0  q_r ^(36) =(q_r ^(13) )^2 q_r ^(10) =(q_r ^4 +2)^2 q_r ^(10) =(q_r ^8 +4q_r ^4 +4)q_r ^(10)   =q_r ^(18) +4q_r ^(14) +4q_r ^(10)   =(q_r ^4 +2)q_r ^5 +4(q_r ^4 +2)q_r +4q_r ^(10)   ⇒Σ_(r=1) ^(13) A_r q_r ^(36) =0  q_r ^(40) =(q_r ^(13) )^3 q_r =(q_r ^4 +2)^3 q_r =(q_r ^(12) +6q_r ^8 +12q_r ^4 +8)q_r   =q_r ^(13) +6q_r ^9 +12q_r ^5 +8q_r    ⇒Σ_(r=1) ^(13) A_r q_r ^(40) =Σ_(r=1) ^(13) A_r q_r ^(13) =2  q_r ^(44) =(q_r ^(13) )^3 q_r ^5 =(q_r ^4 +2)^3 q_r ^5 =(q_r ^(12) +6q_r ^8 +12q_r ^4 +8)q_r ^5    =q_r ^(17) +6q_r ^(13) +12q_r ^9 +8q_r ^5   =(q_r ^4 +2)q_r ^4 +6q_r ^(13) +12q_r ^9 +8q_r ^5   ⇒Σ_(r=1) ^(13) A_r q_r ^(44) =6Σ_(r=1) ^(13) A_r q_r ^(13) =6×2=12  x_(143) =2^(11) +C_(10) ^(11) 2^1 (Σ_(r=1) ^(13) A_r q_r ^(40) )+C_(11) ^(11) 2^0 (Σ_(r=1) ^(13) A_r q_r ^(44) )  x_(143) =2^(11) +11×2×2+1×2^0 ×12  ⇒x_(143) =2^(11) +44+12=2104    in this way we can calculate every  term of the sequence even though  we don′t know the values of A_r  and  q_r  (r=1,2,...,13).

givenistherecursiverelationxn+13=xn+4+2xnweseexn=Aqnfulfillsthiscondition.Aqn+13=Aqn+4+2Aqnq13=q4+2ithas13roots,sayqr(r=1,2,...,13).thegeneralformulafortermxnisthenxn=13r=1Arqrnwithqr13=qr4+2(r=1,2,3,...,13)x0=13r=1Ar=1(x13=x4+2x0x0=1)xm=13r=1Arqrm=0(m=1,2,3,...,12)x13=13r=1Arqr13=2x143=13r=1Arqr143qr143=(qr13)11=(qr4+2)11=11k=0Ck11qr4k211kx143=13r=1Ar(11k=0Ck11qr4k211k)x143=11k=0Ck11211k(13r=1Arqr4k)k=013r=1Arqr4k=1k=1..313r=1Arqr4k=0x134=211+11k=4Ck11211k(13r=1Arqr4k)k=4..11:qr16=qr13qr3=(qr4+2)qr3=qr7+2qr313r=1Arqr16=13r=1Arqr7+213r=1Arqr3=0qr20=qr13qr7=(qr4+2)qr7=qr11+2qr713r=1Arqr20=13r=1Arqr11+213r=1Arqr7=0qr24=qr13qr11=(qr4+2)qr11=(qr4+2)qr2+2qr1113r=1Arqr24=0qr28=(qr13)2qr2=(qr4+2)2qr2=qr10+4qr6+4qr213r=1Arqr28=0qr32=(qr13)2qr6=(qr4+2)2qr6=qr14+4qr10+4qr6=(qr4+2)qr+4qr10+4qr613r=1Arqr32=0qr36=(qr13)2qr10=(qr4+2)2qr10=(qr8+4qr4+4)qr10=qr18+4qr14+4qr10=(qr4+2)qr5+4(qr4+2)qr+4qr1013r=1Arqr36=0qr40=(qr13)3qr=(qr4+2)3qr=(qr12+6qr8+12qr4+8)qr=qr13+6qr9+12qr5+8qr13r=1Arqr40=13r=1Arqr13=2qr44=(qr13)3qr5=(qr4+2)3qr5=(qr12+6qr8+12qr4+8)qr5=qr17+6qr13+12qr9+8qr5=(qr4+2)qr4+6qr13+12qr9+8qr513r=1Arqr44=613r=1Arqr13=6×2=12x143=211+C101121(13r=1Arqr40)+C111120(13r=1Arqr44)x143=211+11×2×2+1×20×12x143=211+44+12=2104inthiswaywecancalculateeverytermofthesequenceeventhoughwedontknowthevaluesofArandqr(r=1,2,...,13).

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