数列{an}满足:1/a1+2/a2+3/a3+…+n/an=2n(1)求通项公式an(2)设Sn=a1+a2+…+an,求1/S1+2/S2+…+n/Sn
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数列{an}满足:1/a1+2/a2+3/a3+…+n/an=2n(1)求通项公式an(2)设Sn=a1+a2+…+an,求1/S1+2/S2+…+n/Sn数列{an}满足:1/a1+2/a2+3/a
数列{an}满足:1/a1+2/a2+3/a3+…+n/an=2n(1)求通项公式an(2)设Sn=a1+a2+…+an,求1/S1+2/S2+…+n/Sn
数列{an}满足:1/a1+2/a2+3/a3+…+n/an=2n
(1)求通项公式an
(2)设Sn=a1+a2+…+an,求1/S1+2/S2+…+n/Sn
数列{an}满足:1/a1+2/a2+3/a3+…+n/an=2n(1)求通项公式an(2)设Sn=a1+a2+…+an,求1/S1+2/S2+…+n/Sn
1,1/a1+2/a2+3/a3+…+n/an=2n
那么1/a1+2/a2+3/a3+…+(n-1)/a(n-1)=2(n-1)
两式相减,得:n/an=2n-2(n-1)=2
那么an=n/2
2,Sn=1/2+2/2+3/2+……+n/2
=(1+2+3+……+n)/2
=n(n+1)/4
那么n/Sn=4/(n+1),
所以1/S1+2/S2+…+n/Sn=4[1/2+1/3+1/4+……+1/(n+1)]
后面好像不能计算了……………………………………
(1)an=n/2
(2)原式=4(1—1/(n+1))
第一问用错位相减
第二问先把sn求出来再列项求和
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