题目
用裂项法求值
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
提问时间：2021-03-31
答案
1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
=[和1/n(n+1)-和1/(n+1)(n+2)]/2
=[1-1/(n+1)-1/2+1/(n+2)]/2
=1/4-1/2(n+1)(n+2)
1/n(n+2)=[1/n-1/(n+2)]/2
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-1/2(n+1)-1/2(n+2)
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
=[和1/n(n+1)-和1/(n+1)(n+2)]/2
=[1-1/(n+1)-1/2+1/(n+2)]/2
=1/4-1/2(n+1)(n+2)
1/n(n+2)=[1/n-1/(n+2)]/2
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-1/2(n+1)-1/2(n+2)
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