《分数裂项求和方法总结(最新整理)》由会员分享，可在线阅读，更多相关《分数裂项求和方法总结(最新整理)（7页珍藏版）》请在金锄头文库上搜索。
1、分数裂项求和方法总结（一）用裂项法求1n(n +1)型分数求和分析：因为1 -1nn +1n +1n(n +1)- n=n(n +1)1n(n +1)（n 为自然数）所以有裂项公式：1= 1 -1 n(n +1)nn +1（二）用裂项法求1型分数求和n(n + k )1分析：型。（n,k 均为自然数）n(n + k )1 111n + kn1因为 (-) = - =k nn + kk n(n + k )n(n + k )n(n + k )1= 1 ( 1 -1 ) 所以 n(n + k )k nn + k（三）用裂项法求k n(n + k )型分数求和k分析：型（n,k 均为自然数）n(n +
2、 k )1 -1 n + k-nk nn + kn(n + k )n(n + k )n(n + k )k所以n(n + k )1 -1nn + k（四）用裂项法求2kn(n + k )(n + 2k )型分数求和2k分析：（n,k 均为自然数）n(n + k )(n + 2k )2k=1-1n(n + k )(n + 2k )n(n + k )(n + k )(n + 2k )（五）用裂项法求1n(n + k )(n + 2k )(n + 3k )型分数求和1分析：（n,k 均为自然数）n(n + k )(n + 2k )(n + 3k )1= 1 (1-1) n(n + k )(n + 2k
3、 )(n + 3k )3k n(n + k )(n + 2k )(n + k )(n + 2k )(n + 3k )（六） 用裂项法求3kn(n + k )(n + 2k )(n + 3k )型分数求和3k分析：（n,k 均为自然数）n(n + k )(n + 2k )(n + 3k )3k=1-1 n(n + k )(n + 2k )(n + 3k )n(n + k )(n + 2k )(n + k )(n + 2k )(n + 3k )记忆方法：1. 看分数分子是否为 1；2. 是 1 时，裂项之后需要整体首尾之差分之一；3. 不是 1 时不用再乘；裂项时首尾各领一队分之一相减。4.“”“
4、”At the end, Xiao Bian gives you a passage. Minand once said, people who learn to learn are very happy people. In every wonderful life, learning is an eternal theme. As a professional clerical and teaching position, I understand the importance of continuous learning, life is diligent, nothing can be
5、 gained, only continuous learning can achieve better self. Only by constantly learning and mastering the latest relevant knowledge, can employees from all walks of life keep up with the pace of enterprise development and innovate to meet the needs of the market. This document is also edited by my studio professionals, there may be errors in the document, if there are errors, please correct, thank you!
