Question

已知正数数列｛an｝满足：2√sn=an+1 （1）求a1，a2，a3，a4 （2）求数列an通项

（3）设bn=1/（an·an+1），求bn前n项和

Answer 1

4Sn=(An+1)^2
=>
4S(n-1)=(A(n-1)+1)^2
=>
4An=(An+1)^2-(A(n-1)+1)^2
=>
(A(n-1)+1)^2=(An-1)^2
=>
A(n-1)+1=An-1 or A(n-1)+1=-An+1
=>
A(n-1)+2=An or A(n-1)=-An
正数列
=>
A(n-1)+2=An
=》
等差，d=2
2√sn=an+1
=>
2√a1=a1+1
=>
a1=1
=>
An=1+(n-1)*2=n+1
=>
Bn
=1/((n+1)(n+2))
=1/(n+1)-1/(n+2)
=>
Sn(b)
=1/2-1/3+1/3-1/4+...+1/(n+1)-1/(n+2)
=1/2-1/(n+2)

Answer 2

2