2025年裂项相消法试题及答案

2025年裂项相消法试题及答案本文借鉴了近年相关经典试题创作而成，力求帮助考生深入理解测试题型，掌握答题技巧，提升应试能力。一、选择题（每题3分，共30分）1.下列哪个级数适合使用裂项相消法求和？A.\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)B.\(\sum_{n=1}^{\infty}\frac{1}{n^2}\)C.\(\sum_{n=1}^{\infty}\frac{1}{n}\)D.\(\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\)2.对于级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)，裂项后相消，最终结果为：A.1B.\(\frac{1}{2}\)C.\(\frac{1}{3}\)D.03.级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)的求和结果为：A.\(\frac{\pi}{4}\)B.\(\frac{1}{2}\)C.1D.\(\frac{1}{4}\)4.若级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\)使用裂项相消法，裂项后的形式为：A.\(\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)B.\(\sum_{n=1}^{\infty}\left(\frac{1}{n(n+1)}-\frac{1}{n+2}\right)\)C.\(\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}\right)\)D.\(\sum_{n=1}^{\infty}\left(\frac{1}{n(n+1)}-\frac{1}{n(n+2)}\right)\)5.级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\)的求和结果为：A.\(\frac{1}{4}\)B.\(\frac{1}{3}\)C.\(\frac{1}{2}\)D.16.级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)裂项后相消，最终结果为：A.\(\frac{\pi}{4}\)B.\(\frac{1}{2}\)C.1D.\(\frac{1}{4}\)7.对于级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)，裂项后相消，最终结果为：A.1B.\(\frac{1}{2}\)C.\(\frac{1}{3}\)D.08.级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)的裂项形式为：A.\(\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)B.\(\sum_{n=1}^{\infty}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)\)C.\(\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}+\frac{1}{2n+1}\right)\)D.\(\sum_{n=1}^{\infty}\left(\frac{1}{2n}+\frac{1}{2n+1}\right)\)9.级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\)裂项后相消，最终结果为：A.\(\frac{1}{4}\)B.\(\frac{1}{3}\)C.\(\frac{1}{2}\)D.110.级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)的求和结果为：A.\(\frac{\pi}{4}\)B.\(\frac{1}{2}\)C.1D.\(\frac{1}{4}\)二、填空题（每题4分，共20分）1.级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)的求和结果为______。2.级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)的裂项形式为______。3.级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\)的求和结果为______。4.级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)的求和结果为______。5.级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)裂项后相消，最终结果为______。三、解答题（每题10分，共50分）1.求级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)的和。2.求级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)的和。3.求级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\)的和。4.求级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)的和。5.求级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)裂项后相消，最终结果。答案和解析一、选择题1.A-裂项相消法适用于分母可以分解为两个连续整数乘积的级数。-\(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)，适合裂项相消法。2.A-\(\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\cdots=1\)3.B-\(\frac{1}{(2n-1)(2n+1)}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)-\(\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots\right)=\frac{1}{2}\)4.D-\(\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}\right)\)5.A-\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}\right)\)-\(\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\cdots\right)=\frac{1}{4}\)6.B-\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\)7.B-\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots=\frac{1}{2}\)8.A-\(\frac{1}{(2n-1)(2n+1)}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)9.A-\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}\right)=\frac{1}{4}\)10.B-\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\)二、填空题1.\(\frac{1}{2}\)-\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2}\)2.\(\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)3.\(\frac{1}{4}\)-\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}\right)=\frac{1}{4}\)4.\(\frac{1}{2}\)-\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\)5.\(\frac{1}{2}\)-\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2}\)三、解答题1.求级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)的和。-\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)-这是一个典型的裂项相消级数，相邻项相消后，只剩下第一项和最后一项的极限：\[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\cdots=1\]-因此，级数的和为\(\frac{1}{2}\)。2.求级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)的和。-\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)-这是一个典型的裂项相消级数，相邻项相消后，只剩下第一项和最后一项的极限：\[\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots\right)=\frac{1}{2}\]-因此，级数的和为\(\frac{1}{2}\)。3.求级数\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\)的和。-\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}\right)\)-这是一个典型的裂项相消级数，相邻项相消后，只剩下第一项和最后一项的极限：\[\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\cdots\right)=\frac{1}{4}\]-因此，级数的和为\(\frac{1}{4}\)。4.求级数\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}\)的和。-\(\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)-这是一个典型的裂项相消级数，相邻项相消后，只剩下第一项和最后一项的极限：\[\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots\right)=\frac{1}{2}\]-因此，级数的和为\(\frac{1