裂項(xiàng)求和試題及答案

一、單項(xiàng)選擇題（每題2分，共10題）1.數(shù)列\(zhòng)(\{a_n\}\)的通項(xiàng)公式\(a_n=\frac{1}{n(n+1)}\)，裂項(xiàng)后為（）A.\(\frac{1}{n}-\frac{1}{n+1}\)B.\(\frac{1}{n+1}-\frac{1}{n}\)C.\(\frac{1}{n}+\frac{1}{n+1}\)D.\(\frac{1}{n-1}-\frac{1}{n}\)2.數(shù)列\(zhòng)(\{a_n\}\)通項(xiàng)\(a_n=\frac{1}{(2n-1)(2n+1)}\)裂項(xiàng)結(jié)果是（）A.\(\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})\)B.\(\frac{1}{2}(\frac{1}{2n+1}-\frac{1}{2n-1})\)C.\(\frac{1}{2n-1}-\frac{1}{2n+1}\)D.\(\frac{1}{2n+1}-\frac{1}{2n-1}\)3.數(shù)列\(zhòng)(\{a_n\}\)中\(zhòng)(a_n=\frac{1}{n(n+2)}\)，裂項(xiàng)為（）A.\(\frac{1}{2}(\frac{1}{n}-\frac{1}{n+2})\)B.\(\frac{1}{n}-\frac{1}{n+2}\)C.\(\frac{1}{2}(\frac{1}{n+2}-\frac{1}{n})\)D.\(\frac{1}{n+2}-\frac{1}{n}\)4.已知\(a_n=\frac{1}{n(n+3)}\)，則\(a_1\)裂項(xiàng)后是（）A.\(\frac{1}{3}(1-\frac{1}{4})\)B.\(\frac{1}{3}(\frac{1}{4}-1)\)C.\(1-\frac{1}{4}\)D.\(\frac{1}{4}-1\)5.數(shù)列\(zhòng)(\{a_n\}\)通項(xiàng)\(a_n=\frac{1}{(n+1)(n+2)}\)，裂項(xiàng)為（）A.\(\frac{1}{n+1}-\frac{1}{n+2}\)B.\(\frac{1}{n+2}-\frac{1}{n+1}\)C.\(\frac{1}{n}-\frac{1}{n+3}\)D.\(\frac{1}{n+3}-\frac{1}{n}\)6.對(duì)于數(shù)列\(zhòng)(\{a_n\}\)，\(a_n=\frac{1}{n(n+4)}\)裂項(xiàng)正確的是（）A.\(\frac{1}{4}(\frac{1}{n}-\frac{1}{n+4})\)B.\(\frac{1}{4}(\frac{1}{n+4}-\frac{1}{n})\)C.\(\frac{1}{n}-\frac{1}{n+4}\)D.\(\frac{1}{n+4}-\frac{1}{n}\)7.數(shù)列\(zhòng)(\{a_n\}\)中\(zhòng)(a_n=\frac{1}{(3n-2)(3n+1)}\)裂項(xiàng)為（）A.\(\frac{1}{3}(\frac{1}{3n-2}-\frac{1}{3n+1})\)B.\(\frac{1}{3}(\frac{1}{3n+1}-\frac{1}{3n-2})\)C.\(\frac{1}{3n-2}-\frac{1}{3n+1}\)D.\(\frac{1}{3n+1}-\frac{1}{3n-2}\)8.已知\(a_n=\frac{1}{n(n+5)}\)，裂項(xiàng)得（）A.\(\frac{1}{5}(\frac{1}{n}-\frac{1}{n+5})\)B.\(\frac{1}{5}(\frac{1}{n+5}-\frac{1}{n})\)C.\(\frac{1}{n}-\frac{1}{n+5}\)D.\(\frac{1}{n+5}-\frac{1}{n}\)9.數(shù)列\(zhòng)(\{a_n\}\)通項(xiàng)\(a_n=\frac{1}{(2n+1)(2n+3)}\)裂項(xiàng)后是（）A.\(\frac{1}{2}(\frac{1}{2n+1}-\frac{1}{2n+3})\)B.\(\frac{1}{2}(\frac{1}{2n+3}-\frac{1}{2n+1})\)C.\(\frac{1}{2n+1}-\frac{1}{2n+3}\)D.\(\frac{1}{2n+3}-\frac{1}{2n+1}\)10.若\(a_n=\frac{1}{n(n+6)}\)，裂項(xiàng)形式為（）A.\(\frac{1}{6}(\frac{1}{n}-\frac{1}{n+6})\)B.\(\frac{1}{6}(\frac{1}{n+6}-\frac{1}{n})\)C.\(\frac{1}{n}-\frac{1}{n+6}\)D.\(\frac{1}{n+6}-\frac{1}{n}\)二、多項(xiàng)選擇題（每題2分，共10題）1.以下哪些數(shù)列通項(xiàng)可以用裂項(xiàng)求和法（）A.\(a_n=\frac{1}{n(n+3)}\)B.\(a_n=\frac{1}{(2n-1)(2n+1)}\)C.\(a_n=n^2\)D.\(a_n=\frac{1}{n(n+1)(n+2)}\)2.數(shù)列\(zhòng)(\{a_n\}\)通項(xiàng)\(a_n=\frac{1}{n(n+2)}\)，裂項(xiàng)求和時(shí)用到的裂項(xiàng)方法有（）A.寫成\(\frac{1}{2}(\frac{1}{n}-\frac{1}{n+2})\)B.消去中間項(xiàng)C.求首項(xiàng)D.確定項(xiàng)數(shù)3.對(duì)于裂項(xiàng)求和，以下說法正確的是（）A.可以將數(shù)列通項(xiàng)拆分成兩項(xiàng)差的形式B.拆分后的項(xiàng)能前后抵消很多項(xiàng)C.只適用于等差數(shù)列D.能簡化數(shù)列求和運(yùn)算4.數(shù)列\(zhòng)(\{a_n\}\)中\(zhòng)(a_n=\frac{1}{(n+1)(n+3)}\)，裂項(xiàng)求和過程包含（）A.裂項(xiàng)為\(\frac{1}{2}(\frac{1}{n+1}-\frac{1}{n+3})\)B.列出前\(n\)項(xiàng)和表達(dá)式C.計(jì)算首項(xiàng)和末項(xiàng)D.化簡求和結(jié)果5.已知\(a_n=\frac{1}{n(n+4)}\)，裂項(xiàng)求和的步驟有（）A.裂項(xiàng)成\(\frac{1}{4}(\frac{1}{n}-\frac{1}{n+4})\)B.求\(S_n\)時(shí)確定哪些項(xiàng)保留C.分析項(xiàng)數(shù)D.得出\(S_n\)的最終表達(dá)式6.以下數(shù)列通項(xiàng)裂項(xiàng)正確的有（）A.\(a_n=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)B.\(a_n=\frac{1}{(2n-1)(2n+1)}=\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})\)C.\(a_n=\frac{1}{n(n+2)}=\frac{1}{2}(\frac{1}{n}-\frac{1}{n+2})\)D.\(a_n=\frac{1}{(n+1)(n+2)}=\frac{1}{n+1}-\frac{1}{n+2}\)7.裂項(xiàng)求和時(shí)，可能涉及到的運(yùn)算有（）A.分式的通分B.項(xiàng)的合并C.確定抵消項(xiàng)D.計(jì)算首末項(xiàng)8.數(shù)列\(zhòng)(\{a_n\}\)通項(xiàng)\(a_n=\frac{1}{(n+2)(n+4)}\)，裂項(xiàng)求和正確做法有（）A.裂項(xiàng)為\(\frac{1}{2}(\frac{1}{n+2}-\frac{1}{n+4})\)B.計(jì)算\(S_n\)時(shí)分析前后項(xiàng)抵消情況C.寫出\(S_n\)的展開式D.化簡\(S_n\)得到最終結(jié)果9.對(duì)于數(shù)列\(zhòng)(\{a_n\}\)，若\(a_n=\frac{1}{n(n+5)}\)，裂項(xiàng)求和的要點(diǎn)包括（）A.正確裂項(xiàng)為\(\frac{1}{5}(\frac{1}{n}-\frac{1}{n+5})\)B.清楚\(n\)項(xiàng)和中哪些項(xiàng)不能抵消C.求出\(S_n\)D.分析數(shù)列的單調(diào)性10.以下關(guān)于裂項(xiàng)求和與數(shù)列的關(guān)系，正確的是（）A.可以解決一些非等差等比數(shù)列的求和問題B.對(duì)通項(xiàng)公式有一定要求C.是一種常見的數(shù)列求和方法D.能直接判斷數(shù)列的通項(xiàng)公式三、判斷題（每題2分，共10題）1.數(shù)列\(zhòng)(\{a_n\}\)通項(xiàng)\(a_n=\frac{1}{n(n+1)}\)，裂項(xiàng)后可寫成\(\frac{1}{n}-\frac{1}{n+1}\)。（）2.裂項(xiàng)求和只適用于分母為兩項(xiàng)乘積形式的數(shù)列通項(xiàng)。（）3.若\(a_n=\frac{1}{(2n-1)(2n+1)}\)，裂項(xiàng)為\(\frac{1}{2}(\frac{1}{2n+1}-\frac{1}{2n-1})\)。（）4.數(shù)列\(zhòng)(\{a_n\}\)中\(zhòng)(a_n=\frac{1}{n(n+2)}\)裂項(xiàng)求和時(shí)，中間項(xiàng)都能消去。（）5.裂項(xiàng)求和過程中，不需要考慮項(xiàng)數(shù)。（）6.已知\(a_n=\frac{1}{(n+1)(n+3)}\)，裂項(xiàng)為\(\frac{1}{2}(\frac{1}{n+1}-\frac{1}{n+3})\)。（）7.所有數(shù)列都能用裂項(xiàng)求和法求前\(n\)項(xiàng)和。（）8.數(shù)列\(zhòng)(\{a_n\}\)通項(xiàng)\(a_n=\frac{1}{n(n+4)}\)裂項(xiàng)后求\(S_n\)，只需保留首項(xiàng)和末項(xiàng)。（）9.裂項(xiàng)求和時(shí)，拆分后的兩項(xiàng)差形式中，分子一定是1。（）10.若\(a_n=\frac{1}{(2n+1)(2n+3)}\)，裂項(xiàng)為\(\frac{1}{2}(\frac{1}{2n+1}-\frac{1}{2n+3})\)。（）四、簡答題（每題5分，共4題）1.簡述裂項(xiàng)求和法的基本原理。裂項(xiàng)求和法是將數(shù)列通項(xiàng)拆分成兩項(xiàng)差的形式，在求前\(n\)項(xiàng)和時(shí)，中間的很多項(xiàng)可以相互抵消，從而簡化求和過程，求出數(shù)列的前\(n\)項(xiàng)和。2.數(shù)列\(zhòng)(\{a_n\}\)通項(xiàng)\(a_n=\frac{1}{n(n+1)}\)，求其前\(n\)項(xiàng)和\(S_n\)。\(a_n=\frac{1}{n}-\frac{1}{n+1}\)，\(S_n=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots+(\frac{1}{n}-\frac{1}{n+1})=1-\frac{1}{n+1}=\frac{n}{n+1}\)。3.數(shù)列\(zhòng)(\{a_n\}\)通項(xiàng)\(a_n=\frac{1}{(2n-1)(2n+1)}\)，裂項(xiàng)后的形式是什么？裂項(xiàng)后為\(\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})\)。4.說明裂項(xiàng)求和法適用的數(shù)列通項(xiàng)特點(diǎn)。適用通項(xiàng)為分式形式，且分母能分解為兩個(gè)因式乘積，兩因式的差為常數(shù)，通過裂項(xiàng)能使前后項(xiàng)相互抵消一部分的數(shù)列。五、討論題（每題5分，共4題）1.討論裂項(xiàng)求和法在數(shù)列求和中的優(yōu)勢和局限性。優(yōu)勢：能解決一些非等差等比數(shù)列求和問題，簡化計(jì)算過程。局限性：對(duì)數(shù)列通項(xiàng)形式要求高，并非所有數(shù)列都適用；裂項(xiàng)時(shí)需要一定技巧，容易出錯(cuò)；對(duì)于復(fù)雜通項(xiàng)，裂項(xiàng)難度大。2.舉例說明如何根據(jù)數(shù)列通項(xiàng)選擇合適的裂項(xiàng)方法。如\(a_n=\frac{1}{n(n+1)}\)，分母是兩個(gè)連續(xù)自然數(shù)乘積，裂項(xiàng)為\(\frac{1}{n}-\frac{1}{n+1}\)；\(a_n=\frac{1}{(2n-1)(2n+1)}\)，分母兩因式差為2，裂項(xiàng)為\(\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})\)，根據(jù)分母特點(diǎn)選擇裂項(xiàng)。3.當(dāng)數(shù)列通項(xiàng)分母有三個(gè)因式時(shí)，如何嘗試裂項(xiàng)求和？可先將通項(xiàng)寫成部分分式形式，例如\(a_n=\frac{1}{n(n+1)(n+2)}=\frac{1}{2}[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}]\)，再分別對(duì)每一項(xiàng)按常規(guī)裂項(xiàng)方法處理，最后求前\(n\)項(xiàng)和。4.探討裂項(xiàng)求和法與其他數(shù)列求和方法的聯(lián)系與區(qū)別。聯(lián)系：都是求數(shù)列前\(n\)項(xiàng)和的方法，有時(shí)可結(jié)合使用。區(qū)別：裂項(xiàng)求和針