为何(2n+1) (2n--1)分之一等于2分之1乘以（2n--1分之1--2n+1分之1）（详解

1/(2n-1)-1/(2n+1)
=[(2n+1)-(2n-1)]/[(2n+1)(2n-1)].
=2*1/[(2n+1)(2n-1)].
<=>1/[(2n+1)(2n-1)]=1/2*[1/(2n-1)-1/(2n+1)].

用代入法

待定系数法
设1/(2n+1)(2n-1)=a/(2n-1)+b/(2n+1)

1/(2n+1) (2n-1)
=(n+1/2-n+1/2)/(2n+1) (2n-1)
=[(n+1/2)-(n-1/2)]/(2n+1) (2n-1)
=(n+1/2)/(2n+1) (2n-1)-(n-1/2)/(2n+1) (2n-1)
=(1/2)(2n+1)/(2n+1) (2n-1)-(1/2)(2n-1)/(2n+1) (2n-1)
=(1/2)*1/(2n-1)-(1/2)*1/(2n+1)
=(1/2)*[1/(2n-1)-1/(2n+1)]

1/((2n+1)(2n-1))=((2n+1)-(2n-1))/(2(2n+1)(2n-1))=1/2*(1/(2n-1)-1/(2n+1))

1 2 (2n+1)-(2n-1) 1 1 1
------------=-------------=---------------=---(----- - -----)
(2n+1)(2n-1) 2(2n+1)(2n-1) 2(2n+1)(2n-1) 2 2n-1 2n+1

这样写能看懂吗?赫赫