已知数列an前N项和为sn,点（n,sn)都在函数f(x)=2x^2-x上,设bn=sn/(n+p),且数列bn是等差数

sn＝2n^2－n,bn＝sn/(n+p)＝(2n^2-n)/(n+p)
b1＝1/(1+p),b2=6/(2+p),b3=15/(3+p).
bn是等差数列,则b1+b3=2b2,即1/(1+p)＋15/(3+p)＝12/(p+2),通分,解得p＝0