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)!

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.