Question

若a1=1/2,an=a1+2a2+3a3.....+(n-1)a(n-1) n>1,求an

Answer 1

楼上的回答错了吧。

an = a1 + 2a2 + ... + (n-2)a(n-2) + (n-1)a(n-1)
= a(n-1) + (n-1)a(n-1)
= n * a(n-1)
= n * (n-1) * a(n-2)
= ...
= n * (n-1) * (n-2) * ... * 3 * 2 * a1
= n! * a1 = n!/2

n=1时也满足，所以这就是所求的an

希望有用，谢谢采纳 ^_^

Answer 2

解：a2=a1=1/2
a(n-1)=a1+2a2+3a3+……+(n-2)a(n-2)
an=a1+2a2+3a3+……+(n-2)a(n-2)+(n-1)a(n-1)
=a(n-1)+(n-1)a(n-1)
=na(n-1)
=n*(n-1)a(n-2)
=……=n*(n-1)*(n-2)*……*3a2
=n!/4

Answer 3

an=a1+2a2+3a3.....+(n-1)a(n-1)
a(n+1)=a1+2a2+3a3.....+(n-1)a(n-1)+nan
a(n+1)-an=nan a(n+1)= (n+1)an a(n+1)/ an =(n+1) an /a(n-1)=n
an /a(n-1)* a(n-1)/a(n-2)* * a2/a1=n*(n-I )*(n-2)……2＝n！
an ＝a1*n！＝n！/4