已知正数数列an的前n项和为sn,满足sn^2=a1^3+.an^3.（1）求证an为等差数列,并求出通项公式

1、∵Sn^2=a1^3+a2^3+…+an^3,
∴Sn-1^2=a1^3+a2^3+…+a(n-1)^3,
两式相减,得an^3=Sn^2-S(n-1)^2=（Sn-S(n-1)）)（Sn+S(n-1)）)=an（Sn+S(n-1)）,
∵an＞0,∴an^2=Sn+S(n-1)（n≥2）,
∴a(n-1 )^2=S(n-1)+S(n-2()n≥2),
两式相减,得an2-an-12 =Sn-S(n-2)=an+a(n-1),
∴an-a(n-1)=1（n＞3）,
∵S1^2=a1^2=a1^3,且a1＞0,∴a1=1,
S2^2=(a1+a2)^2=a1^3+a2^3,
∴（1+a2）^2=1+a2^3,∴a2^3-a2^2-2a2=0,
由a2＞0,得a2=2,
∴an-a(n-1)=1,n≥2,
故数列{an}为等差数列,通项公式为an=n．
2、bn=(1-1/n)^2-a(1-1/n)^2=1/n^2+(a-2)/n+1-a,
b(n+1)-bn=(1/(n+1)-1/n)(1/(n+1)+1/n+a-2)=-[1/n(n+1)][1/(n+1)+1/n+a-2]>0
即1/(n+1)+1/n+a-2)