Question

已知数列{an}的前n项和为Sn,a1=1,Sn=n/n+2an+1则an=_.

Answer 1

解: ∵数列{an}的前n项和为Sn,∴an=Sn-Sn-1,n>1,∵Sn=n/n+2an+1,∴Sn-1=n-1/n+1an则 Sn-Sn-1=n/n+2an+1-n-1/则耐袭n+1an,即an=n/n+2an+1-n-1/n+1an,∴an+1/an=2(n+2)/n+1(n>1),∵a1=1,S1=1/3a2,∴a2=3,∴a2/a1=3/1,∵a3/a2=2×4/3,a4/a3=2×5/4,a5/a4=2×6/5,...,an/an-1=2(n+1)/n,累乘得an/a1=(n+1).2n-2,an=(n+1).2n-2,当n=1时适合 综上孙兄亩哗所述,答案:(n+1).2n-2