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

Question

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

Answer 1

(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)] +1
S(n+1)/[(n+1)(n+2)]-Sn/[n(n+1)]=1,为定值
S1/(1×2)=a1/(1×2)=3/2,数列{Sn/[n(n+1)]}是以3/2为首项,1为公差的等差数列
Sn/[n(n+1)]=3/2 +1×(n-1)=(2n+1)/2
Sn=n(n+1)(2n+1)/2
n≥2时,an=Sn-S(n-1)=n(n+1)(2n+1)/2 - n(n-1)(2n-1)/2=3n²
n=1时,a1=3×1²=3,同样满足通项公式
数列{an}的通项公式为an=3n²
2.
考察一般项：12/ak=12/(3k²)=4/k²
12/a1+12/a2+12/a3+12/a4+...+12/an
=4/(1/1²+1/2²+1/3²+1/4²...+1/n²)
0 4/n>0 7-4/n