(1)设等差数列{an}首项为a1,公差为d,由题意得 a1+2d=2 15a1+ 15×14 2 d=105 ,…3分解得a1=0,d=1,∴an=n-1.…6分(2)bn=3 an+2n=3n-1+2n,…7分∴Tn=(30+31+32+…+3n-1)+2(1+2+3+…+n)= 1?3n 1?3 +n(n+1)= 3n 2 +n(n+1)? 1 2 .…12分