Question

11. 已知数列{an}的前n项和为Sn,a1=1,且an= 2Sn2Sn- 1 (n≥2),求数列{an}的通项公

Answer 1

an=2sn2sn-1
sn-sn-1=2sn2sn-1 两边同除snsn-1得
1/(sn-1)-1/sn=4 所以1/sn为公差为-4的等差数列
1/sn=1/s1+(n-1)*(-4) 因为1/s1=1/a1=1
1/sn=5-4n
sn=1/(5-4n)
an=sn-sn-1=1/(5-4n)-1/(5-4(n-1))
an=[(9-4n)-(5-4n)]/[(5-4n)(9-4n)]
an=4/[(5-4n)(9-4n)] (n=>2)