(2014•商丘三模)设等差数列{an}的前n项和为Sn,已知(a2012-1)3+2014a2012=0,(a3-1)3+2014a3=4028,则下列结论正确的是( )
A. S2014=2014,a2012<a3
B. S2014=2014,a2012>a3
C. S2014=2013,a2012<a3
D. S2014=2013,a2012>a3
人气:401 ℃ 时间:2020-01-30 10:22:45
解答
构造函数f(x)=(x-1)3+2014x,则f′(x)=3(x-1)2+2014>0,∴函数f(x)=(x-1)3+2014x单调递增,∵f(a3)=4028>f(a2012)=0,∴a2012<a3,排除B和D,已知两式相加可得(a2012-1)3+2014a2012+(a3-1)3+...
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