(1)由已知f(1)=n2+2n,可得a1+a2+…+an=n2+2n,令,当n=1时,a1=S1;当n≥2时,an=Sn-Sn-1即可得出
(2)利用(1)即可得出..
【解析】
(1)由已知f(1)=n2+2n,可得a1+a2+…+an=n2+2n,令,
∴当n=1时,a1=1+2=3;
当n≥2时,an=Sn-Sn-1=n2+2n-[(n-1)2+2(n-1)]
=2n+1,n=1时也成立.
∴an=2n+1.
(2)由(1)可得an=2n+1.
∴f(3)=3×3+5×32+7×33=117.