Question

(n-1)n(n+1)分之一怎么裂项？

Answer 1

若是为了分解成单项式,楼上3位高手都给出答案啦.
若是为了裂项求和.则如下~
1/[(n-1)n(n+1)] = (1/2)/[(n-1)n] - (1/2)/[n(n+1)]
这样,
1/(1*2*3) + 1/(2*3*4) + ... + 1/[(n-1)n(n+1)]
=(1/2){1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+...+1/[(n-1)n]-1/[n(n+1)]}
= (1/2){1/(1*2)-1/[n(n+1)]}
=1/4 - (1/2)/[n(n+1)]

Answer 2

1/2（1/（n一1）一2/n十1/（n十1））

Answer 3

A/(n-1)+B/n+C/(n+1) =1/(n-1)n(n+1)
同乘(n-1)n(n+1)
An(n+1)+B(n-1)(n+1)+C(n-1)n=1
先令n=-1, C*(-1-1)(-1)=1,C=1/2
n=0,B(-1)(1)=1,B=-1
n=1,A(1)(2)=1,A=1/2
1/(n-1)n(n+1)=1/2(n-1)-1/n+1/2(n+1)

Answer 4

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