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21.已知函數(shù)f(x)=x3+ax2+bx+c關(guān)于點(1,1)成中心對稱,且f '(1)=0.
(Ⅰ)求函數(shù)f(x)的表達式;
(Ⅱ)設(shè)數(shù)列{an}滿足條件:a1∈(1,2),an+1=f (an)
求證:(a1- a2).(a3-1)+(a2- a3).(a4-1)+…+(an- an+1).(an+2-1)<1
解:(Ⅰ)由f(x)=x3+ax2+bx+c關(guān)于點(1,1)成中心對稱,所以
x3+ax2+bx+c+(2-x)3+a(2-x)2+b(2-x)+c=2
對一切實數(shù)x恒成立.得:a=-3,b+c=3,
對由f '(1)=0,得b=3,c=0,
故所求的表達式為:f(x)= x3-3x2+3x.
(Ⅱ) an+1=f (an)= an 3-3 an 2+3 an (1)
令bn=an-1,0<bn<1,由代入(1)得:bn+1=,bn=,
∴ 1>bn >bn+1 >0
(a1-a2).(a3-1)+(a2-a3).(a4-1)+…+(an-an+1).(an+2-1)=
<=b1-bn+1<b1<1?! ?