已知数列An满足An＞0,其前n项和为Sn为满足2Sn=An的平方+An(1)求An(2)设数列Bn满足An/2的n次方,Tn=b1+b2+

题目

已知数列An满足An＞0,其前n项和为Sn为满足2Sn=An的平方+An(1)求An(2)设数列Bn满足An/2的n次方,Tn=b1+b2+

答案

（1）2Sn=an^2+an2Sn-1=a(n-1)^2+a(n-1)2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)an^2-a(n-1)^2=an+a(n-1)[an+a(n-1)][an-a(n-1)]=an+a(n-1)an>0 an+a(n-1)>0,等式两边同除以an+a(n-1)an-a(n-1)=1an=a(n-1)+12a1=2S1=a...