a1=1,a2=6,sn=3sn-1-2sn-2+2^n(n大于等于3)求证:an/2^n是等差数列
已知等差数列{an}的前N项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n大于等于2)
已知等差数列{an}的前N项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n大于等于2),
已知数列an中,a1=1,当n大于等于2时,sn=an(1-2/sn).求证1/sn是等差数列
在数列{an}中,有a1=3,Sn=a1+a2+...+an,2an=Sn*S(n+1)(n大于等于2)
已知a1=3,an=Sn-1+2^n(n大于等于2),求an,Sn?
数列{an}中,a1=1,an=2Snˆ2/(2Sn-1)(n大于等于2) 1.证明{1∕Sn}是等差数列 2
设Sn是数列an的前n项和,已知a1=1,an=-Sn*Sn-1,(n大于等于2),则Sn=
已知数列{an}的前n项和为Sn,且满足Sn=Sn-1/2Sn-1 +1,a1=2,求证{1/Sn}是等差数列
已知等差数列{an}的前N项和为Sn,a1=-2/3,满足Sn+1/Sn+2=an(n大于等于2) 求Sn 别用数学归纳
已知数列an首相a1=3,通项an和前n项和SN之间满足2an=Sn*Sn-1(n大于等于2)
a1=1,n,an,Sn成等差数列,证明{Sn+n+2}是等比数列
数列an中,a1=1,当n大于=2时,sn满足sn方=an(sn-1) 证明1/sn是等差数列