∴2S=(2+1n1-(2-19-12n+12n-1一=2,n-1令bn=2n+1,则b-b=2(n≥2)1∴数列{}是以-=二=1为首项,公差d=2的等差∴b=2-1,2i+12-1S通车∴SS2S=3x-xx+1=2m+1,2-1由2m+12019,得m1009,即正整数m的最小信为1009,故应1009