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)!
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an+1an=n+1(n≥2),又a1=1,∴a2=1,∴a3a2=3,a4a3=4,…,anan−1=n.累积得an=n!2...
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
=n+1(n≥2),累乘即可得到答案,注意n的范围.an+1 an
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.