Question

已知数列an的前n项和为Sn,并且满足a1=2,n*an+1=Sn+n(n+1).

(1)求an的通项公式
(2)令Tn=(4/5)^*Sn,问是否存在正整数m,对一切正整数n,总有Tn小于等于Tm,若存在,求m的值:若不存在,说明理由.

Answer 1

n*a(n+1)=S(n)+n(n+1).
(n+1)a(n+2)=S(n+1)+(n+1)(n+2)
a(n+1)=S(n+1)-S(n)=(n+1)a(n+2)-(n+1)(n+2)-n*a(n+1)+n(n+1)
=(n+1)a(n+2)-na(n+1)-2(n+1)
a(n+2)=a(n+1)+2
a(n)=2n S(n)=n(n+1)
Tn=(4/5)^Sn=(4/5)^[n(n+1)]随n增大而下降
m=1即可