最美时光续文:这个题目结果是什么?

来源:百度文库 编辑:查人人中国名人网 时间:2024/05/05 09:12:51
题目是 算 (2/3*4/3)*(4/5*6/5)*。。。。*(2n/2n+1*2n+2/2n+1)

(2/3*4/3)*(4/5*6/5)*....*(2n/2n+1*2n+2/2n+1)
=4/2*6/4*8/6*...*(2n+2)/2n
=(2n+2)/2
=n+1

(2/3*4/3)*(4/5*6/5)*....*(2n/2n+1*2n+2/2n+1)
=4/2*6/4*8/6*...*(2n+2)/2n
=(2n+2)/2
=n+1

2/3*4/3
是指(2/3)*(4/3)还是(2)/(4*3)/3 ?

(2/3*4/3)*(4/5*6/5)*…*[2n/(2n+1)*(2n+2)/(2n+1)]
={2/3*4/5*6/7*…*[2n/(2n+1)]}*{4/3*6/5*8/7*…*[(2n+2)/(2n+1)]}
=2^(2n)*{1/3*2/5*3/7*…*[n/(2n+1)]}*{2/3*3/5*4/7*…*[(n+1)/(2n+1)]}

1/3*2/5*3/7*…*[n/(2n+1)]
分子、分母同乘以2*4*6*8…2n,即(2^n)*(n!),上式变为
n!*(2^n)*(n!)/(2n+1)!=(2^n)*(n!)^2/(2n+1)!

同理:
2/3*3/5*4/7*…*[(n+1)/(2n+1)]可以简化为
(n+1)!*(2^n)*(n!)/(2n+1)!=(2^n)*(n!)^2*(n+1)/(2n+1)!

所以
原式=2^(2n)*[(2^n)*(n!)^2/(2n+1)!]*[(2^n)*(n!)^2*(n+1)/(2n+1)!]
=2^(4n)*(n!)^4*(n+1)/[(2n+1)!]^2

(2/3*4/3)*(4/5*6/5)*....*(2n/2n+1*2n+2/2n+1)
=4/2*6/4*8/6*...*(2n+2)/2n
=(2n+2)/2
=n+1
只是我的想法,不清楚对不对!!!!!