Question

数列an的前n项和为Sn,已知a1=1,an+1=(n+2)Sn/n(n=1,2...

数列an的前n项和为Sn,已知a1=1,an+1=(n+2)Sn/n(n=1,2,3····),证明数列{Sn/n} 数列an的前n项和为Sn,已知a1=1,an+1=(n+2)Sn/n（n=1,2,3····）,证明数列{Sn/n}是等比数列

Answer 1

由a(n+1)=(n+2)Sn/n
而a(n+1)=S(n+1)-Sn
所以S(n+1)-Sn=(n+2)Sn/n
化为S(n+1)/(n+1)=2Sn/n
则数列{Sn/n}是首项为1,公比为2的等比数列
所以Sn/n=2^(n-1)