裂項(xiàng)考試題及答案

單項(xiàng)選擇題1.式子$\frac{1}{1\times2}+\frac{1}{2\times3}$裂項(xiàng)后是（）A.$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}$B.$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}$C.$1-\frac{1}{2}-\frac{1}{2}-\frac{1}{3}$D.$1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}$2.$\frac{1}{n(n+1)}$裂項(xiàng)結(jié)果為（）A.$\frac{1}{n}-\frac{1}{n+1}$B.$\frac{1}{n}+\frac{1}{n+1}$C.$\frac{1}{n}-\frac{1}{n-1}$D.$\frac{1}{n+1}-\frac{1}{n}$3.$\frac{1}{2\times4}$裂項(xiàng)為（）A.$\frac{1}{2}-\frac{1}{4}$B.$\frac{1}{2}\times(\frac{1}{2}-\frac{1}{4})$C.$\frac{1}{2}\times(\frac{1}{2}+\frac{1}{4})$D.$\frac{1}{2}+\frac{1}{4}$4.算式$\frac{1}{3\times5}+\frac{1}{5\times7}$裂項(xiàng)后為（）A.$\frac{1}{2}\times(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7})$B.$\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}$C.$\frac{1}{2}\times(\frac{1}{3}-\frac{1}{5}-\frac{1}{5}-\frac{1}{7})$D.$\frac{1}{3}-\frac{1}{5}-\frac{1}{5}-\frac{1}{7}$5.$\frac{2}{n(n+2)}$裂項(xiàng)結(jié)果是（）A.$\frac{1}{n}-\frac{1}{n+2}$B.$\frac{1}{n}+\frac{1}{n+2}$C.$2\times(\frac{1}{n}-\frac{1}{n+2})$D.$2\times(\frac{1}{n}+\frac{1}{n+2})$6.$\frac{1}{4\times6}$裂項(xiàng)為（）A.$\frac{1}{4}-\frac{1}{6}$B.$\frac{1}{2}\times(\frac{1}{4}-\frac{1}{6})$C.$\frac{1}{2}\times(\frac{1}{4}+\frac{1}{6})$D.$\frac{1}{4}+\frac{1}{6}$7.$\frac{1}{1\times3}+\frac{1}{3\times5}$裂項(xiàng)后是（）A.$\frac{1}{2}\times(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5})$B.$1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}$C.$\frac{1}{2}\times(1+\frac{1}{3}-\frac{1}{3}-\frac{1}{5})$D.$1+\frac{1}{3}-\frac{1}{3}-\frac{1}{5}$8.$\frac{1}{5\times7}$裂項(xiàng)結(jié)果為（）A.$\frac{1}{5}-\frac{1}{7}$B.$\frac{1}{2}\times(\frac{1}{5}-\frac{1}{7})$C.$\frac{1}{2}\times(\frac{1}{5}+\frac{1}{7})$D.$\frac{1}{5}+\frac{1}{7}$9.對于$\frac{1}{6\times8}$，裂項(xiàng)正確的是（）A.$\frac{1}{6}-\frac{1}{8}$B.$\frac{1}{2}\times(\frac{1}{6}-\frac{1}{8})$C.$\frac{1}{2}\times(\frac{1}{6}+\frac{1}{8})$D.$\frac{1}{6}+\frac{1}{8}$10.$\frac{3}{n(n+3)}$裂項(xiàng)為（）A.$\frac{1}{n}-\frac{1}{n+3}$B.$\frac{1}{n}+\frac{1}{n+3}$C.$3\times(\frac{1}{n}-\frac{1}{n+3})$D.$3\times(\frac{1}{n}+\frac{1}{n+3})$多項(xiàng)選擇題1.可裂項(xiàng)的式子有（）A.$\frac{1}{2\times3}$B.$\frac{1}{3\times5}$C.$\frac{1}{n(n+1)}$D.$\frac{1}{m(m+2)}$2.下列裂項(xiàng)結(jié)果正確的有（）A.$\frac{1}{3\times4}=\frac{1}{3}-\frac{1}{4}$B.$\frac{1}{4\times6}=\frac{1}{2}\times(\frac{1}{4}-\frac{1}{6})$C.$\frac{2}{5\times7}=2\times(\frac{1}{5}-\frac{1}{7})$D.$\frac{1}{n(n+3)}=\frac{1}{3}\times(\frac{1}{n}-\frac{1}{n+3})$3.關(guān)于$\frac{1}{n(n+k)}$（$k$為常數(shù)）裂項(xiàng)說法正確的是（）A.裂項(xiàng)結(jié)果為$\frac{1}{k}\times(\frac{1}{n}-\frac{1}{n+k})$B.當(dāng)$k=1$時(shí)，裂項(xiàng)為$\frac{1}{n}-\frac{1}{n+1}$C.裂項(xiàng)有助于簡便計(jì)算D.裂項(xiàng)結(jié)果可以是$\frac{1}{n}+\frac{1}{n+k}$4.能將式子裂項(xiàng)后進(jìn)行簡便計(jì)算的是（）A.$\sum_{n=1}^{10}\frac{1}{n(n+1)}$B.$\sum_{n=1}^{5}\frac{1}{n(n+2)}$C.$\sum_{n=3}^{7}\frac{1}{(n-1)(n+1)}$D.$\sum_{n=1}^{20}\frac{1}{n(n+3)}$5.已知式子$\frac{1}{a(a+b)}$（$a,b$為正數(shù)），其裂項(xiàng)相關(guān)內(nèi)容正確的是（）A.裂項(xiàng)后為$\frac{1}\times(\frac{1}{a}-\frac{1}{a+b})$B.當(dāng)$b=2$時(shí)，裂項(xiàng)是$\frac{1}{2}\times(\frac{1}{a}-\frac{1}{a+2})$C.裂項(xiàng)能使求和更復(fù)雜D.結(jié)合裂項(xiàng)可將原式寫成$\frac{1}{a}-\frac{1}{a+b}$的形式6.以下屬于裂項(xiàng)形式應(yīng)用的有（）A.計(jì)算$\frac{1}{2\times3}+\frac{1}{3\times4}$B.化簡$\sum_{n=2}^{6}\frac{1}{n(n-1)}$C.求$\frac{1}{1\times5}$的值D.估算$\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{99\times100}$的范圍7.對于式子$\frac{1}{(x+1)(x+3)}$，下列說法正確的是（）A.可裂項(xiàng)為$\frac{1}{2}\times(\frac{1}{x+1}-\frac{1}{x+3})$B.裂項(xiàng)后方便計(jì)算它的和C.它不能進(jìn)行裂項(xiàng)D.若求和$\sum_{x=1}^{4}\frac{1}{(x+1)(x+3)}$，裂項(xiàng)后可簡便運(yùn)算8.裂項(xiàng)法在（）方面有作用。A.簡化分?jǐn)?shù)求和B.證明等式C.求解數(shù)列問題D.計(jì)算復(fù)雜根式9.式子$\frac{1}{(y-2)(y-4)}$（$y\neq2,y\neq4$）裂項(xiàng)相關(guān)正確的有（）A.裂項(xiàng)后為$\frac{1}{2}\times(\frac{1}{y-4}-\frac{1}{y-2})$B.裂項(xiàng)后利于計(jì)算$\sum_{y=5}^{8}\frac{1}{(y-2)(y-4)}$C.若直接求和會更簡便D.裂項(xiàng)時(shí)要保證分母不為零10.下列式子裂項(xiàng)正確的有（）A.$\frac{1}{5\times8}=\frac{1}{3}\times(\frac{1}{5}-\frac{1}{8})$B.$\frac{1}{7\times10}=\frac{1}{3}\times(\frac{1}{7}-\frac{1}{10})$C.$\frac{1}{9\times12}=\frac{1}{3}\times(\frac{1}{9}-\frac{1}{12})$D.$\frac{1}{11\times14}=\frac{1}{3}\times(\frac{1}{11}-\frac{1}{14})$判斷題1.$\frac{1}{3\times4}$裂項(xiàng)結(jié)果是$\frac{1}{3}+\frac{1}{4}$。（）2.裂項(xiàng)法只能用于分?jǐn)?shù)求和。（）3.$\frac{1}{n(n+2)}$裂項(xiàng)為$\frac{1}{2}\times(\frac{1}{n}-\frac{1}{n+2})$。（）4.裂項(xiàng)后計(jì)算不一定會更簡便。（）5.$\frac{1}{5\times7}$裂項(xiàng)是$\frac{1}{5}-\frac{1}{7}$。（）6.對于式子$\frac{1}{(m-1)m}$，裂項(xiàng)結(jié)果是$\frac{1}{m-1}-\frac{1}{m}$。（）7.裂項(xiàng)法在所有數(shù)列求和中都適用。（）8.$\frac{1}{6\times9}$裂項(xiàng)為$\frac{1}{3}\times(\frac{1}{6}-\frac{1}{9})$。（）9.裂項(xiàng)后中間項(xiàng)一定能全部抵消。（）10.$\frac{1}{(x+2)(x+4)}$裂項(xiàng)為$\frac{1}{2}\times(\frac{1}{x+2}-\frac{1}{x+4})$。（）簡答題1.簡述裂項(xiàng)法的原理。2.寫出$\frac{1}{4\times7}$的裂項(xiàng)形式并說明理由。3.裂項(xiàng)法在分?jǐn)?shù)求和中有什么優(yōu)勢？4.對于$\frac{1}{n(n+4)}$如何進(jìn)行裂項(xiàng)？討論題1.討論裂項(xiàng)法在不同類型數(shù)列求和中的適用性。2.當(dāng)裂項(xiàng)后中間項(xiàng)不能完全抵消時(shí)，該如何處理？3.結(jié)合實(shí)際例子，討論裂項(xiàng)法對簡化計(jì)算的作用。4.探討裂項(xiàng)法在證明等式方面可能的應(yīng)用。答案單項(xiàng)選擇題1.A2.A3.B4.A5.A6.B7.A8.B9.B10.A多項(xiàng)選擇題1.ABCD2.ABD3.ABC4.ABCD5.AB6.ABD7.ABD8.ABC9.ABD10.ABCD判斷題1.×2.×3.√4.√5.×6.√7.×8.√9.×10.√簡答題1.裂項(xiàng)法原理是把數(shù)列的通項(xiàng)拆分成兩項(xiàng)之差，在求和時(shí)使大部分中間項(xiàng)相互抵消，從而達(dá)到簡化計(jì)算的目的。2.$\frac{1}{4\times7}=\frac{1}{3}\times(\frac{1}{4}-\frac{1}{7})$，因?yàn)?\frac{1}{n(n+k)}=\frac{1}{k}\times(\frac{1}{n}-\frac{1}{n+k})$，這里$n=4$，$k=3$。3.裂項(xiàng)法可將復(fù)雜分?jǐn)?shù)求和轉(zhuǎn)化為可抵消的形式，減少計(jì)算量，使求和過程更簡單快捷。4.$\frac{1}{n(n+4)}=\frac{1}{4}\times(\frac{1}{n}-\frac{1}{n+4})$，依據(jù)裂項(xiàng)公式$\frac{1}{n(n+k)}=\frac{1}{k}\times(\frac{1}{n}-\frac{1}{n+k})$，$k=4$。討論題1.裂項(xiàng)法適用于通項(xiàng)可拆成兩項(xiàng)差形式的數(shù)列，如分式數(shù)列。但對于無