数列{an}满足a1=a,an+1=can-c(n属于N*),a,c为实数,c不等于0,求数列{an}的通项公式

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数列{an}满足a1=a,an+1=can-c(n属于N*),a,c为实数,c不等于0,求数列{an}的通项公式
数列{an}满足a1=a,an+1=can-c(n属于N*),a,c为实数,c不等于0,求数列{an}的通项公式

数列{an}满足a1=a,an+1=can-c(n属于N*),a,c为实数,c不等于0,求数列{an}的通项公式
a(n+1)=can-c
a2=ca1-c=ac-c
an=ca(n-1)-c
a(n+1)-an=c[an-a(n-1)]
设bn=a(n+1)-an,b1=a2-a1=ac-c-a
bn=cb(n-1)
bn=b1*c^(n-1)
=(ac-c-a)*c^(n-1)
a(n+1)-an=(ac-c-a)*c^(n-1)
an-a(n-1)=(ac-c-a)*c^(n-2)
a(n-1)-a(n-2)=(ac-c-a)*c^(n-3)
……
a4-a3=(ac-c-a)*c^2
a3-a2=(ac-c-a)*c^1
a2-a1=(ac-c-a)*c^0
两边相加：
an-a1=(ac-c-a)[c^(n-2)+c^(n-3)+c^(n-3)+……+c^3+c^2+c^1+1]
c=1时,
an-a1=(ac-c-a)(n-1)
an=a+(ac-c-a)(n-1)
c≠1时,
an-a1=(ac-c-a)[1-c^(n-1)]/(1-c)
an=a1+(ac-c-a)[1-c^(n-1)]/(1-c)
=a+(ac-c-a)[1-c^(n-1)]/(1-c)
=c/(c-1)+[(ac-c-a)/(c-1)]c^(n-1)

An+1-1=c(An-1)
当a=1时，An=1
当a≠1时，An=(a-1)*c^(n-1)+1

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