已知数列{an}的前N项和为Sn,a1=1.nan+1=(n+2)Sn

题目

已知数列{an}的前N项和为Sn,a1=1.nan+1=(n+2)Sn
nan+1的n+1是下角标哈.
求证数列{Sn/n}为等比数列
求an的通项公式及前N项和Sn

答案

nan+1=(n+2)Sn,则
(n-1)an=(n+1)S(n-1)
相减,
nan+1/(n+2)-(n-1)an/(n+1)=an
即
nan+1/(n+2)=2nan/(n+1)
则,
an+1/an=2(n+2)/(n+1)
则,
(Sn/n)/(Sn/(n-1))=(an+1/(n+2))/(an/(n+1))
=(n+1)/(n+2)*2(n+2)/(n+1)
=2
则{Sn/n}为等比数列
Sn/n=(1-2^n)/(1-2)=2^n-1
则,Sn=n(2^n-1)
则S(n-1)=(n-1)(2^(n-1)-1)
相减,
an=n(2^n-1)-（n-1)(2^(n-1)-1)
得
an=(n+1)*2^(n-1)+1