已知数列{an},a1=1,前n项和为Sn,且点P(an,an+1)(n∈N*)在直线x-y+1=0上,则1S1+1S2
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已知数列{an},a1=1,前n项和为Sn,且点P(an,an+1)(n∈N*)在直线x-y+1=0上,则
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∵点P(an,an+1)(n∈N*)在直线x-y+1=0上
∴an-an+1+1=0 ∴数列{an}是以1为首项,以1为公差的等差数列. ∴an=n ∴sn= n(n+1) 2 ∴ 1 sn= 2 n(n+1)=2( 1 n− 1 n+1) 1 S1+ 1 S2+ 1 S3+…+ 1 Sn=2(1− 1 2+ 1 2− 1 3+…+ 1 n− 1 n+1)= 2n n+1 故选C
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