已知数列{an}满足a1=1,an=a1+2a2+3a3+…+(n-1)an-1,则n≥2时,数列{an}的通项an=(  )A. n!2B. (n+1)!2C. n!D. (n+1)!

问题描述:

已知数列{an}满足a1=1,an=a1+2a2+3a3+…+(n-1)an-1,则n≥2时,数列{an}的通项an=(  )
A.

n!
2

B.
(n+1)!
2

C. n!
D. (n+1)!

由an=a1+2a2+3a3+…+(n-1)an-1(n≥2),得
nan+an=a1+2a2+3a3+…+(n-1)an-1+nan(n≥2),
∴(n+1)•an=an+1(n≥2),则

an+1
an
=n+1(n≥2),
又a1=1,∴a2=1,
a3
a2
=3,
a4
a3
=4,…,
an
an−1
=n.
累积得an=
n!
2
(n≥2),
故选A.
答案解析:由an=a1+2a2+3a3+…+(n-1)an-1(n≥2),得nan+an=a1+2a2+3a3+…+(n-1)an-1+nan(n≥2),整理可得
an+1
an
=n+1
(n≥2),累乘即可得到答案,注意n的范围.
考试点:数列递推式.
知识点:本题考查由数列递推式求数列的通项,累乘法是求数列通项公式的常用方法,要准确把握其解决方法及使用条件.