- 数列问题4已知数列{an}满足a1=1,an=a1+2a2+3a
- 已知数列{an}满足a1=1,an=a1+2a2+3a3+……+(n-1)a(n-1)(n≥2)
则{an}的通项an=?
- an = a1 + 2a2 + 3a3 + … + (n-1)a(n-1) (n≥2)
a(n+1) = a1 + 2a2 + 3a3 + … + (n-1)a(n-1) + nan (n≥1)
相减得 a(n+1) - an = nan (n≥2)
即 a(n+1) = (n+1)an (n≥2)
所以 an = na(n-1)
a(n-1) = (n-1)a(n-2)
………………………
a3 = 3a2
a2 = 1a1
a1 = 1
相乘得 an = 1*1*3*…*n = 1*2*3*…*n / 2 = n!/2 (n≥2时)
另当n=1时,an = 1
(最后这个符号“!”认识吗?——就是1,2,3...到n的连乘积)