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
问题描述:
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
已知函数f(x)=(x-1)^2,g(x)=4(x-1),数列An满足a1=2,且(a(n+1)-an)g(an)+f(an)=0
(注意其中n和n+1为下标) .bn=3/4(n+2)(an-1)(n-1不是下标 n是),若t^m/b^m
答
(a(n+1)-an)4(an-1)+(an-1)²=0(an-1)(4a(n+1)-4an+an-1)=04a(n+1)=3an+1a(n+1)-1=(3/4)(an-1)an-1为首项为1公比为3/4的等比数列2.an-1=(3/4)^(n-1)an=(3/4)^(n-1)+1sn=(1-(3/4)^n)/(1-3/4)+nsn=4-4*(3/4)^n+n≥...