已知数列{an}中，a1=1，前n项和Sn=[(n+2)/3]an．
(1)求a2，a3；
(2)求{an}的通项公式．
(1)由a1=1与Sn=[(n+2)/3]an 可得 S2=[(2+2)/3]a2=a1+a2 a2⇒3a1=3 S3=[(3+2)/3]a3=a1+a1+a2 ⇒ (2/3)a3= a1+a2=4 ⇒ a3=6 故所求a2,a3的值分别为3，6． (2)当n≥2时，Sn=[(n+2)/3]an①，Sn-1=[(n+1)/3]an-1②，①-②可得Sn-Sn-2=[(n+2)/3]an-[(n+1)/3]an-1,既an=[(n+2)/3]an-[(n+1)/3]an-1 ⇔[(n-1)/3]an=[(n+1)/3]an-1=(n+1)/(n-1) 故有an=[an/an-1]×[an-1/an-2]×…×[a2/a1]×a1=(n+1)/(n-1)×[n/(n/2)]×…×3/1×1=(n2+n)/2. 而(1 2+1)/2=1=a 1，所以{an}的通项公式为an=(n2+n)/2．
