已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于正数)在直线x-y+1=0上则1/S1+1/S2+.+1/S已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于正数)在直线x-y+1=0上则1/S1+1/S2+.......+1/Sn=
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已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于正数)在直线x-y+1=0上则1/S1+1/S2+.+1/S已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于正数)在直线x-y+1=0上则1/S1+1/S2+.......+1/Sn=
已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于正数)在直线x-y+1=0上则1/S1+1/S2+.+1/S
已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于正数)在直线x-y+1=0上则1/S1+1/S2+.......+1/Sn=
已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于正数)在直线x-y+1=0上则1/S1+1/S2+.+1/S已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于正数)在直线x-y+1=0上则1/S1+1/S2+.......+1/Sn=
将点带进直线方程 得数列an是首项为1公差为1的等差数列 即an通项公式为an=n
所以Sn=(n+1)n/2 1/Sn=2/n(n+1)
所以1/S1+1/S2+……+1/Sn=2【1/1*2+1/2*3+……+1/n(n+1)】=2(1-1/n+1)=2n/n+1
点P(an,an+1)在直线x-y+1=0上 则带入有:an-a(n+1)+1=0 即a(n+1) =an+1为公差为一的等差数列 a1=1 an=n sn=n(n+1)/2 1/sn=2/n(n+1)=2(1/n-1/n+1)
所求式子=2(1-1/2+1/2-1/3+……+1/n-1/n+1)=1-1/n+1=n/n+1