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
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![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](/uploads/image/z/303452-44-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%A6%96%E9%A1%B9a1%3D3%2F5%2Can%2B1%3D3an%2F2an%2B1%2Cn%3D1%2C2%2C%E2%80%A6.%2C%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F2an%2B1)
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
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
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
(3的n次方)除以(2+3的n次方)
是2an+1还是2(an+1)啊
解答如下图
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)...
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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)
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