Question

已知数列{an}的首项a1=3/5,an+1=3an/2an+1, n=1,2,…. ,求{an}的通项公式

2an+1

Answer 1

解答如下图

Answer 2

解答如下图

Answer 3

an+1=3an/(2an+1)
所以:an+1=3an/(2an+1)
=3*(3a(n-1)/((2a(n-1)+1))/(2*((3a(n-1)/((2a(n-1)+1))+1)
=9a(n-1)/(8a(n-1)+1)
=27a(n-2)/(26a(n-2)+1)
...
=(3^((n+1)-1)a1/((3^((n+1)-1)-1)a1+1)
=(3^n)a1/(((3^n)-1)a1+1)
=(3^(n+1))/(3((3^n)-1)+5)

an=(3^(n))/(3((3^(n-1))-1)+5)

(验证:由上式可得:a1=3/5 (满足已知条件),a2=9/11
将它们代入:an+1=3an/2an+1,也满足)

Answer 4

（3的n次方）除以（2+3的n次方）

Answer 5

是2an+1还是2（an+1）啊