若Sn是数列an的前n项和,a1=3,2Sn=na(n+1)-n(n+1)(n+2)
问题描述：
若Sn是数列an的前n项和,a1=3,2Sn=na(n+1)-n(n+1)(n+2)
（1）求数列an的通项公式
（2）证明：对一切正整数n,有12/a1+12/a2+.+12/an＜7
答
(1)2Sn=na(n+1)-n(n+1)(n+2)2Sn=n[S(n+1)-Sn]-n(n+1)(n+2)nS(n+1)=(n+2)Sn+n(n+1)(n+2)等式两边同除以n(n+1)(n+2)S(n+1)/[(n+1)(n+2)]=Sn/[n(n+1)] +1S(n+1)/[(n+1)(n+2)]-Sn/[n(n+1)]=1,为定值S1/(1×2)=a1/(1×2)=...