Question

已知正项数列{an}的前n项和为Sn,且Sn=1/4(an+1)².(n∈N*).(1)求a1.a2；(2)求

Answer 1

Sn=(1/4)(an+1)^2

n=1

4a1=(a1+1)^2

a1=2

for n>=2

an = Sn -S(n-1)

4an =(an+1)^2 - (a(n-1)+1)^2

(an)^2 - [a(n-1)]^2 - 2[an+ a(n-1) ] =0

[an+ a(n-1) ].[an- a(n-1)-2 ]=0

an- a(n-1)-2

an -a1 =2(n-1)

an =2n

{an}是等差数列

a1=2,

a2=4