Question

已知函数f(x)=(x-1)^2,g(x)=4(x-1),数列An满足a1=2,且(an+1-an)g(an)+f(an)=0

(1)用an表示an＋1；
(2)证明数列{ an－1}是等比数列；

Answer 1

(1)
(a(n+1)-an)g(an)+f(an)=0
(a(n+1)-an)[4(an -1)]+(an -1)^2=0
(4a(n+1).an - 4a(n+1) - 4(an)^2 + 4an) +( (an)^2 -2an +1) =0
4a(n+1) . ( an -1 ) = 3(an)^2-2an-1
= (3an+1)(an-1)
a(n+1) = (3an+1)/4
(2)
a(n+1) = (3an+1)/4
a(n+1) -1 = (3/4)( an -1)
=>{ an－1}是等比数列

Answer 2

(a<n+1>-an)g(an)+f(an)
=(a<n+1>-an)*4(an-1)+(an-1)^2
=(an-1)(4a<n+1>-3an-1)
=0,
a1=2,
∴a1-1≠0，
∴4a<n+1>-3an-1=0，
∴a<n+1>-1=(3/4)(an-1),①
（1）a<n+1>=(3/4)an+1/4.
(2)由①，数列{an-1}是首项为1，公比为3/4的等比数列.