题目

数列中的拆项公式
1/(ab)=
1/[n(n+1)]=
1/[(2n-1)(2n+1)]=
1/[n(n+1)(n+2)]=
要详细地计算过程
答案我有，我要详细地计算过程，请大家帮忙

提问时间：2021-03-17

答案

因为 1/a - 1/b = (b-a)/(ab)
所以 1/(ab)= (1/a - 1/b)/(b-a) = (1/a - 1/b)*(1/(b-a))
a=n,b=n+1时:
1/(n(n+1))= 1/n - 1/(n+1)
a=2n-1,b=2n+1时:
1/[(2n-1)(2n+1)] = 1/2 * [1/(2n-1) - 1/(2n+1)]
1/[n(n+1)(n+2)]可以拆成
X/[n(n+1)] - Y/[(n+1)(n+2)]
用待定系数法算出 X=Y=0.5
如果是为了数列求和就不用再拆了,因为A[n]减去的正好是A[n+1]加的.
要拆的话
1/[n(n+1)(n+2)]=0.5*{1/[n(n+1)]-1/[(n+1)(n+2)]}=0.5[1/n-1/(n+1)-1/(n+1)+1/(n+2)]=0.5[1/n -2/(n+1) +1/(n+2)]