Question

已知函数f(x)=-2x+1,当x∈[An,Bn]时,f(x)的值域为[A(n+1),B(n+1)],a1=0.b1=1

Answer 1

f(x)=-2x+1 很显然f(x)是一个单调递减函数

f(An)=B(n+1),即：-2An+1=B(n+1) (1)

f(Bn)=A(n+1),即：-2Bn+1=A(n+1) (2)

(2)-(1)得：

A(n+1)-B(n+1)=-2Bn+2An=2(An-Bn)

所以

{An-Bn}是等比数列,A1-B1=-1 公比q=2

数列{An},{Bn}的前n项和为Sn.Tn

则

Sn-Tn=(-1)(1-2^n)/(1-2)=1-2^n, Tn-Sn=2^n-1

g(n)=(T1+T2+.Tn)-(S1+S2+.Sn)

=(T1-S1)+(T2-S2)+...+(Tn-Sn)

=2-1+2^2-1+...+2^n-1

=2(1-2^n)/(1-2)-n

=2^(n+1)-n-2