设数列an的前n项和为Sn 已知a1+2a2+3a3+……+nan=(n-1)Sn+2n
设数列an的前n项和为Sn 已知a1+2a2+3a3+……+nan=(n-1)Sn+2n
抽去{an}中a1,a4,a7.a(3n-2),.余下的顺序不变组成{bn},若{bn}前n项和为Tn,求证12/5<T(n+1)/Tn≤11/3
抽去{an}中a1,a4,a7.a(3n-2),.余下的顺序不变组成{bn},若{bn}前n项和为Tn,求证12/5<T(n+1)/Tn≤11/3
数学人气:287 ℃时间:2019-09-22 07:40:17
优质解答
由a1+2a2+3a3+……+nan=(n-1)Sn+2n可知:n=1时:a1=(1-1)s1+2,解得:a1=2;n=2时:a1+2a2=(2-1)s2+4,即2+2a2=(2+a2)+4,解得:a2=4.由题意,有:a1+2a2+3a3+……+nan=(n-1)Sn+2na1+2a2+3a3+……+nan+(n+1)a(n+1)=[(n+1...
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