Question

已知正项数列〔an〕的前n项和为Sn，且4Sn＝(an＋1)²，怎么求数列的通项公式

已知正项数列〔an〕的前n项和为Sn，且4Sn＝(an＋1)²，怎么求数列的通项公式在线急等，谢谢

Answer 1

4Sn＝(an＋1)^2
n=1
4a1 =(a1＋1)^2
(a1-1)^2 =0
a1=1
for n≥ 2
an = Sn - S(n-1)
4an＝(an＋1)^2 - (a(n-1)＋1)^2
(an)^2 -2an - [a(n-1)]^2 -2a(n-1) =0
[an -a(n-1) ]. [an +a(n-1) ] - 2[an +a(n-1) ] =0
[an +a(n-1) ].[an -a(n-1) -2] =0
an -a(n-1) -2=0
an -a(n-1) =2
=> {an} 是等差数列, d=2
an -a1 =2(n-1)
an -1 = 2n-2
an =2n-1

Answer 2

4Sn-4Sn-1=(an+1)^2-(an-1+1)^2
4an=an^2+2an-an-1^2-2an-1
an^2-an-1^2=2an+2an-1
an-an-1=2
等差d=2
4s1=(a1+1)^2
4a1=(a1+1)^2
a1=1
an=1+2(n-1)
an=2n-1