(1)S1=1、S2=9、S3=36.
T1=1、T2=9、T3=36.
(2)猜Sn=Tn.
用数学归纳法证明:
1)n=1时,Sn=Tn成立.
2)假设n=k时,有Sn=Tn成立,即Sk=Tk.
3)求证:n=k+1时,Sn=Tn也成立,即S(k+1)=T(k+1).
S(k+1)=Sk+(k+1)^3
=Tk+(k+1)^3
=[k(k+1)/2]^2+(k+1)^3
=(k+1)^2[(k/2)^2+k+1]
=(1/4)(k+1)^2(k^2+4k+4)
=(1/4)(k+1)^2(k+2)^2
=[(k+1)(k+2)/2]^2
=T(k+1)
所以,n=k+1时,Sn=Tn成立.
由此可得:Sn=Tn,n为正整数.