数列{an}和{bn}适合下列关系式an=5an-1-6bn-1,bn=3an-1-4bn-1,且a1=a,b1=b,求
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数列{an}和{bn}适合下列关系式an=5an-1-6bn-1,bn=3an-1-4bn-1,且a1=a,b1=b,求通项an和bn.
∵an=5an-1-6bn-1,bn=3an-1-4bn-1,
两式相减得,an-bn=2an-1-bn-1
∴数列{an-bn}为等比数列,公比为2
∴an-bn=(a1-b1)2n-1
=(a-b)2n-1
∴an=bn+(a-b)2n-1
an-1=bn-1+(a-b)2n-2
∴bn+(a-b)2n-1=5[bn-1+(a-b)2n-2)]-6bn-1
bn=-bn-1+3(a-b)2n-2
设cn=bn-(a-b)2n-1,c1=b1-(a-b)=2b-a
cn=-c(n-1)
∴cn=c1(-1)n-1=(2b-a)(-1)n-1
即bn-(a-b)2n-1=cn=(2b-a)(-1)n-1
bn=(a-b)2n-1+(2b-a)(-1)n-1
∴an=bn+(a-b)2n-1=(a-b)2n+(2b-a)(-1)n-1
∴an=(a-b)2n+(2b-a)(-1)n-1
bn=(a-b)2n-1+(2b-a)(-1)n-1
两式相减得,an-bn=2an-1-bn-1
∴数列{an-bn}为等比数列,公比为2
∴an-bn=(a1-b1)2n-1
=(a-b)2n-1
∴an=bn+(a-b)2n-1
an-1=bn-1+(a-b)2n-2
∴bn+(a-b)2n-1=5[bn-1+(a-b)2n-2)]-6bn-1
bn=-bn-1+3(a-b)2n-2
设cn=bn-(a-b)2n-1,c1=b1-(a-b)=2b-a
cn=-c(n-1)
∴cn=c1(-1)n-1=(2b-a)(-1)n-1
即bn-(a-b)2n-1=cn=(2b-a)(-1)n-1
bn=(a-b)2n-1+(2b-a)(-1)n-1
∴an=bn+(a-b)2n-1=(a-b)2n+(2b-a)(-1)n-1
∴an=(a-b)2n+(2b-a)(-1)n-1
bn=(a-b)2n-1+(2b-a)(-1)n-1
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