数列{an}是等差数列,a1=f(x+1),a2=0,a3=f(x-1),其中f(x)=x2-4x+2,则通项公式an=( ) A.-2n+4 B.-2n-4 C.2n-4或-2n+4 D.2n-4
数列{an}是等差数列,a1=f(x+1),a2=0,a3=f(x-1),其中f(x)=x2-4x+2,则通项公式an=( )
A. -2n+4
B. -2n-4
C. 2n-4或-2n+4
D. 2n-4
A. -2n+4
B. -2n-4
C. 2n-4或-2n+4
D. 2n-4
数学人气:845 ℃时间:2020-06-18 07:08:04
优质解答
∵f(x)=x2-4x+2,∴a1=f(x+1)=(x+1)2−4(x+1)+2=x2-2x-1,a3=f(x−1)=(x−1)2−4(x−1)+2=x2-6x+7,又数列{an}是等差数列,a2=0∴a1+a3=2a2=0,∴(x2-2x+1)+(x2-6x+7)=2x2-4x6=0,解得:x=1或x=3 &nb...
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