a1=1,n,an,Sn成等差数列,证明｛Sn+n+2}是等比数列

a1=1,n,an,Sn成等差数列,证明｛Sn+n+2}是等比数列 数列An的前n项和为Sn,已知A1=1,An+1=Sn*(n+2)/n,证明数列Sn/n是等比数列 设等比数列{an}中,a1=256,前n项和为Sn,且Sn,Sn+2,Sn+1成等差数列, 在数列{An}中,已知A1=1,An=2Sn^2/(2Sn-1),(n>=2),证明{1/Sn}是等差数列,并求Sn 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 等比数列的证明方式数列An的前n项和为Sn,A1=1,A(n+1)=2Sn+1,证明数列An是等比数列 数列前n项和为sn,a1=1,an+sn是公差为2的等差数列,求an-2是等比数列,并求sn 已知数列｛an}的前n项和为Sn,a1=1,数列｛an+Sn}是公差为2的等差数列 证明｛an-2}是等比数列 an=n 数列an ,a1=1,当n>=2时,an=（根号sn+根号sn-1）/2,证明根号sn是等差数列,求an 利用等差数列求和公式Sn=n(a1+an)/2证明Sn=na1+n(n-1)/2*d 证明数列是等比数列数列前n项和为Sn,a1=1,a(n+1)=（n+2)Sn/n,求证Sn/n是等比数列, 已知数列｛an｝中,a1=1,且Sn,Sn+1,2S1成等差数列,用数学归纳法证明Sn=(2^n-1)/2^(n-1)