如何用数学归纳法证明？

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)]