All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 81308 by naka3546 last updated on 11/Feb/20
Commented by mr W last updated on 11/Feb/20
2104?
Commented by naka3546 last updated on 11/Feb/20
please,showyourworking,sir.
Answered by mr W last updated on 12/Feb/20
givenistherecursiverelationxn+13=xn+4+2xnweseexn=Aqnfulfillsthiscondition.Aqn+13=Aqn+4+2Aqn⇒q13=q4+2ithas13roots,sayqr(r=1,2,...,13).thegeneralformulafortermxnisthenxn=∑13r=1Arqrnwithqr13=qr4+2(r=1,2,3,...,13)x0=∑13r=1Ar=1(x13=x4+2x0⇒x0=1)xm=∑13r=1Arqrm=0(m=1,2,3,...,12)x13=∑13r=1Arqr13=2x143=∑13r=1Arqr143qr143=(qr13)11=(qr4+2)11=∑11k=0Ck11qr4k211−kx143=∑13r=1Ar(∑11k=0Ck11qr4k211−k)x143=∑11k=0Ck11211−k(∑13r=1Arqr4k)k=0⇒∑13r=1Arqr4k=1k=1..3⇒∑13r=1Arqr4k=0⇒x134=211+∑11k=4Ck11211−k(∑13r=1Arqr4k)k=4..11:qr16=qr13qr3=(qr4+2)qr3=qr7+2qr3⇒∑13r=1Arqr16=∑13r=1Arqr7+2∑13r=1Arqr3=0qr20=qr13qr7=(qr4+2)qr7=qr11+2qr7⇒∑13r=1Arqr20=∑13r=1Arqr11+2∑13r=1Arqr7=0qr24=qr13qr11=(qr4+2)qr11=(qr4+2)qr2+2qr11⇒∑13r=1Arqr24=0qr28=(qr13)2qr2=(qr4+2)2qr2=qr10+4qr6+4qr2⇒∑13r=1Arqr28=0qr32=(qr13)2qr6=(qr4+2)2qr6=qr14+4qr10+4qr6=(qr4+2)qr+4qr10+4qr6⇒∑13r=1Arqr32=0qr36=(qr13)2qr10=(qr4+2)2qr10=(qr8+4qr4+4)qr10=qr18+4qr14+4qr10=(qr4+2)qr5+4(qr4+2)qr+4qr10⇒∑13r=1Arqr36=0qr40=(qr13)3qr=(qr4+2)3qr=(qr12+6qr8+12qr4+8)qr=qr13+6qr9+12qr5+8qr⇒∑13r=1Arqr40=∑13r=1Arqr13=2qr44=(qr13)3qr5=(qr4+2)3qr5=(qr12+6qr8+12qr4+8)qr5=qr17+6qr13+12qr9+8qr5=(qr4+2)qr4+6qr13+12qr9+8qr5⇒∑13r=1Arqr44=6∑13r=1Arqr13=6×2=12x143=211+C101121(∑13r=1Arqr40)+C111120(∑13r=1Arqr44)x143=211+11×2×2+1×20×12⇒x143=211+44+12=2104inthiswaywecancalculateeverytermofthesequenceeventhoughwedon′tknowthevaluesofArandqr(r=1,2,...,13).
Terms of Service
Privacy Policy
Contact: info@tinkutara.com