已知函数f(x)=(x-1)^2,g(x)=10(x-1),数列{an}、{bn}满足 2009-

已知函数f(x)=(x-1)^2,g(x)=10(x-1),数列{an}、{bn}满足a1=2,
∵(an+1-an) g(an)+f(an)=0,
∴[a(n+1)-an]10(an-1)+(an-1)^2=0
∵ an-1不恒为0 ,约去an-1
∴10[a(n+1)-an]=1-an
10a(n+1)=9an+1
a(n+1)=9/10an+1/10
a(n+1)-1=9/10(an-1)
[a(n+1)-1]/(an-1)=9/10
∴数列{an-1}为等比树立,公比为9/10
首项a1-1=1
∴an-1=(9/10)^(n-1)
∴an=1+(9/10)^(n-1)