问题
填空题
[2014·浙江调研]设Sn是数列{an}的前n项和,已知a1=1,an=-Sn·Sn-1(n≥2),则Sn=________.
答案
依题意得Sn-1-Sn=Sn-1·Sn(n≥2),整理得-
=1,又
=
=1,则数列{
}是以1为首项,1为公差的等差数列,因此
=1+(n-1)×1=n,即Sn=
.
[2014·浙江调研]设Sn是数列{an}的前n项和,已知a1=1,an=-Sn·Sn-1(n≥2),则Sn=________.
依题意得Sn-1-Sn=Sn-1·Sn(n≥2),整理得-
=1,又
=
=1,则数列{
}是以1为首项,1为公差的等差数列,因此
=1+(n-1)×1=n,即Sn=
.