高中不等式,有点难!an=1/nSn是an前n项和试比较是前2的n次方和不是Sn
an的前n项和Sn,a1=1,an+1=(n+2)/nSn,证数列Sn/n是等比数列和Sn+1=4an
an的前n项和Sn,a1=7,an+1=(n+2)/nSn,证数列Sn/n是等比数列和S(n+1)=4an
数列an的前n项和为Sn,已知a1=1,an+1=(n+2)/nSn.求证:(1)数列{Sn/n}是等比数列(2)Sn+
已知数列{an}的前n项和为Sn,a1=1,an+1=n+2nSn(n≥1,n∈N*).
设等差数列{an}的前 n项和为Sn,且 Sn=(an+1)^/2(n属于N*)若bn=(-1)nSn,求数列{bn}的
数列{an}的前n项和为Sn,已知a1+2,Sn+1=Sn-2nSn+1Sn,求an
设数列{an}的前n项和为Sn,已知a1=5,且nSn+1=2n(n+1)+(n+1)Sn
已知数列an的前n项和为sn,若a1=1,nsn+1-(n+1)sn=n*n+cn(c是整数,n=1,2,3...)且s
设等差数列{an}的前n项和为Sn,且Sn=((an+1)/2)平方(n属于正整数),若bn=(-1)^nSn,求数列{
设Sn是等比数列{an}的前n项和,且Sn=2an+n
数列的前n项和为Sn,a1=1,an+1=n+2/nSn,求{an}的通项公式
Sn是数列an的前n项和,an=1/n(n+2),求Sn