问题补充:
单选题数列{an}的前n项和,则当n≥2时,有A.Sn>na1>nanB.Sn<nan<na1C.nan>Sn>na1D.nan<Sn<na1
答案:
D解析分析:由已知,求出a1=S1=1,当n≥2时,an=-4n+5.代入化简出表达式,并作差比较即得出正确选项.解答:因为,所以a1=S1=2,当n≥2时,an=Sn-Sn-1=5n-3n2-5(n-1)+3(n-1)2=-6n+8.①Sn-na1 =(5n-3n2)-2n=-3n(n-1)<0,∴Sn<na1 ②na1-nan =6n(n-1)>0,∴na1>nan??? ③Sn-nan=5n-3n2-n(-6n+8)=3n(n-1)>0,∴Sn>nan?? 综合①②③得出na1>Sn>nan.故选D.点评:本题考查数列的通项公式求解及应用.不等式大小比较,考查作差、变形、计算判断能力.