已知数列{an}的前n项的和为Sn=1/4n方+2/3n+3,求这个数列的通项公式.

∵a1=1/4*1²+2/3*1+3=1/4+2/3+3=47/12 an=Sn-Sn-1=1/4n²+2/3n+3-【1/4*（n-1）²+2/3*（n-1）+3】=1/4n²+2/3n+3-【1/4*（n²-2n+1）+2/3n-2/3+3】=1/4n²+2/3n+3-【1/4n²-1/2n+1/4+2/3n-2/3+3】=1/4n²+2/3n+3-【1/4n²+1/6n+31/12】=1/4n²+2/3n+3-1/4n²-1/6n-31/12=1/2n+5/12 (n≥2)当n=1时，a1=S1=1/2*1+5/12=11/12 an=11/12(n=1)或an=1/2n+5/12(≥2）∴数列｛an｝的通式公式为 an=1/2n+5/12 收起