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)!
答
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
=n+1(n≥2),an+1 an
又a1=1,∴a2=1,
∴
=3,a3 a2
=4,…,a4 a3
=n.an an−1
累积得an=
(n≥2),n! 2
故选A.