各项均为正数的数列{an}的前n项和为Sn满足2Sn=an(an+1),n∈N*,求an

(1)
2Sn=an(an+1) (1)
n=1
2a1=a1(a1+1)
a1^2-a1=0
a1=1
2S(n-1)=a(n-1)(a(n-1)+1) (2)
(1)-(2)
2an = (an)^2 +an - [a(n-1)]^2 - a(n-1)
(an)^2 - [a(n-1)]^2 - [an +a(n-1)]=0
[an + a(n-1)].[an-a(n-1) -1] =0
an-a(n-1) -1 =0
an-a(n-1)=1
{an}是等差数列,d=1
an - a1= n-1
an =n
(2)
cn = a(n+1) ; n is odd
= 3.2^(a(n-1)) + 1 ; n is even
c1+c3+...+c(2n-1)
=a2+a4+...+a(2n)
= 2+4+...+2n
=(n+1)n (3)
c2+c4+.+c(2n)
=(3.2^1 + 1)+(3.2^3 + 1)+.+(3.2^(2n-1) + 1)
= n + 2[2^(2n) -1] (4)
T(2n) = c1+c2+.+c(2n)
= (n+1)n +n + 2[2^(2n) -1]
=(n+2)n +2[2^(2n) -1]