an>0
n=1時
S1=a1=(a1²+a1)/2
∴a1=1
n>=2時
S(n-1)=(a(n-1)²-a(n-1))/2
an=Sn-S(n-1)
∴(an+a(n-1))(an-a(n-1)-1)=0
an>0
∴an-a(n-1)=1
∴{an}是等差數(shù)列
an=1+n-1=n
a2=2,a3=3
（3）
(2^n)▪a1▪a2.an≥M√（2n+1）▪（2a1-1）▪（2a2-1）.2(an-1)
∴M≤[(2^n)▪a1▪a2.an]/[√（2n+1）▪（2a1-1）▪（2a2-1）.2(an-1)]
設(shè)f(n)=[(2^n)▪a1▪a2.an]/[√（2n+1）▪（2a1-1）▪（2a2-1）.2(an-1)]
f(n+1)=[(2^(n+1)*1*2*3...*n*(n+1)]/[√(2n+3)*(1*3*5*.(2n-1)(2n+1)]
f(n+1)/f(n)
=(2n+2)/[√(2n+1)*√(2n+3)]
=√[(4n^2+8n+4)/(4n^2+8n+3)]
>1
∴f(n)是增函數(shù)
∴f(n)>=1=2√3/3
∴0