【思路点拨】(1)根据二次函数的导函数为f"(x)=6x-2,可求f(x)=3x2-2x,所以Sn=3n2-2n.由Sn可求an. (2)根据an求cn,求出cn代入2b1+22b2+23b3+…+2nbn=cn中可求出bn,注意n=1与n≥2的讨论. 解:(1)已知二次函数f(x)=px2+qx(p≠0), 则f"(x)=2px+q=6x-2,故p=3,q=-2, 所以f(x)=3x2-2x. 点(n,Sn)(n∈N*)均在函数y=f(x)的图象上, 则Sn=3n2-2n,当n=1时,a1=S1=1; 当n≥2时,an=Sn-Sn-1=6n-5, 故数列{an}的通项公式:an=6n-5. (2)由(1)得,cn=(an+2)=2n-1, 2b1+22b2+23b3+…+2nbn=2n-1, 当n=1时,b1=, 当n≥2时,2b1+22b2+23b3+…+2n-1bn-1+2nbn =2n-1, 2b1+22b2+23b3+…+2n-1bn-1=2(n-1)-1, 两式相减得:bn==21-n, 故数列{bn}的通项公式:bn= |