证明：（1）由题意可得，

an+1

n+1

-

an

n

=2∴{

an

n

}为等差数列．

an

n

=

a1

1

+2(n-1)=2n+1∴an=2n2+n．

（2）由（1）可得，b_n=n•3ⁿ

S_n=1•3+2•3²+…n•3ⁿ

3S_n=1•3²+2•3³+…n•3ⁿ⁺¹

∴-2S_n=3+3²+…+3ⁿ-n•3ⁿ⁺¹=

3(1-3n)

1-3

-n•3n+1

∴Sn=

3+(2n-1)•3n+1

4