累加法經(jīng)典題庫及答案

一、單項選擇題(每題2分,共10題)1.已知數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}-a_n=2\),\(a_1=1\),則\(a_5\)的值為（）A.7B.8C.9D.102.數(shù)列\(zhòng)(\{a_n\}\)中,\(a_1=2\),\(a_{n+1}-a_n=n+1\),則\(a_6\)等于()A.20B.21C.22D.233.若數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}-a_n=3^n\),\(a_1=1\),則\(a_3\)的值為()A.13B.12C.11D.104.已知\(a_1=1\),\(a_{n+1}-a_n=2n\),則\(a_{10}\)的值為()A.81B.82C.90D.915.數(shù)列\(zhòng)(\{a_n\}\)中,\(a_1=3\),\(a_{n+1}-a_n=2n\),則\(a_{10}\)的值是()A.93B.94C.95D.966.已知數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}-a_n=2n-1\),\(a_1=1\),則\(a_4\)的值為()A.7B.8C.9D.107.若\(a_1=2\),\(a_{n+1}-a_n=n+2\),則\(a_5\)的值為()A.15B.16C.17D.188.數(shù)列\(zhòng)(\{a_n\}\)中,\(a_1=0\),\(a_{n+1}-a_n=2^n\),則\(a_6\)的值為()A.62B.63C.64D.659.已知\(a_1=1\),\(a_{n+1}-a_n=3n\),則\(a_4\)的值為()A.13B.14C.15D.1610.數(shù)列\(zhòng)(\{a_n\}\)中,\(a_1=4\),\(a_{n+1}-a_n=3n-2\),則\(a_5\)的值為()A.26B.27C.28D.29二、多項選擇題(每題2分,共10題)1.對于數(shù)列\(zhòng)(\{a_n\}\),若\(a_{n+1}-a_n=f(n)\),以下哪些情況可以用累加法求\(a_n\)（）A.\(f(n)=n\)B.\(f(n)=2^n\)C.\(f(n)=\frac{1}{n(n+1)}\)D.\(f(n)=\sinn\)2.已知數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}-a_n=2n\),\(a_1=1\),則以下說法正確的是()A.\(a_2=3\)B.\(a_3=7\)C.\(a_4=13\)D.\(a_5=21\)3.若數(shù)列\(zhòng)(\{a_n\}\)中,\(a_{n+1}-a_n=3^n\),\(a_1=1\),則()A.\(a_2=4\)B.\(a_3=13\)C.\(a_4=40\)D.\(a_5=121\)4.數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}-a_n=n+1\),\(a_1=1\),則()A.\(a_2=3\)B.\(a_3=6\)C.\(a_4=10\)D.\(a_5=15\)5.已知\(a_{n+1}-a_n=2n-1\),\(a_1=1\),則()A.\(a_2=2\)B.\(a_3=5\)C.\(a_4=10\)D.\(a_5=17\)6.對于數(shù)列\(zhòng)(\{a_n\}\),\(a_{n+1}-a_n=2^n\),\(a_1=1\),以下結(jié)論正確的是()A.\(a_2=3\)B.\(a_3=7\)C.\(a_4=15\)D.\(a_5=31\)7.若\(a_{n+1}-a_n=3n+1\),\(a_1=2\),則()A.\(a_2=6\)B.\(a_3=12\)C.\(a_4=22\)D.\(a_5=36\)8.數(shù)列\(zhòng)(\{a_n\}\)中,\(a_{n+1}-a_n=n^2\),\(a_1=1\),則()A.\(a_2=2\)B.\(a_3=6\)C.\(a_4=15\)D.\(a_5=31\)9.已知\(a_{n+1}-a_n=\frac{1}{n(n+1)}\),\(a_1=1\),則()A.\(a_2=\frac{3}{2}\)B.\(a_3=\frac{5}{3}\)C.\(a_4=\frac{7}{4}\)D.\(a_5=\frac{9}{5}\)10.對于數(shù)列\(zhòng)(\{a_n\}\),\(a_{n+1}-a_n=\cosn\pi\),\(a_1=0\),則()A.\(a_2=-1\)B.\(a_3=0\)C.\(a_4=-1\)D.\(a_5=0\)三、判斷題(每題2分,共10題)1.若數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}-a_n=n\),\(a_1=1\),一定可以用累加法求\(a_n\)。()2.對于數(shù)列\(zhòng)(\{a_n\}\),\(a_{n+1}-a_n=\sinn\),不能用累加法求\(a_n\)。()3.已知\(a_{n+1}-a_n=2^n\),\(a_1=1\),則\(a_2=3\)。()4.若\(a_{n+1}-a_n=3n-1\),\(a_1=2\),用累加法可求\(a_n\)通項公式。()5.數(shù)列\(zhòng)(\{a_n\}\)中,\(a_{n+1}-a_n=\frac{1}{n}\),\(a_1=1\),用累加法求\(a_n\)時能得到簡單的通項公式。()6.已知\(a_{n+1}-a_n=5\),\(a_1=3\),則\(a_3=13\)。()7.若\(a_{n+1}-a_n=n^2\),\(a_1=0\),可以用累加法求\(a_n\)。()8.對于數(shù)列\(zhòng)(\{a_n\}\),\(a_{n+1}-a_n=2n+1\),\(a_1=1\),\(a_5=25\)。()9.已知\(a_{n+1}-a_n=\ln(n+1)\),\(a_1=0\),能用累加法求\(a_n\)。(）10.若\(a_{n+1}-a_n=3^n\),\(a_1=1\),則\(a_4=40\)。（）四、簡答題(每題5分,共4題)1.簡述累加法求數(shù)列通項公式的適用條件。2.已知\(a_1=1\),\(a_{n+1}-a_n=2n\),求\(a_n\)的通項公式。3.若數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}-a_n=3^n\),\(a_1=1\),如何求\(a_n\)?4.用累加法求數(shù)列\(zhòng)(\{a_n\}\)通項公式時,\(a_{n+1}-a_n=f(n)\),\(f(n)\)通常有哪些常見形式？五、討論題(每題5分,共4題)1.討論當(dāng)\(f(n)\)為復(fù)雜函數(shù)時,累加法求數(shù)列通項公式的困難及可能的解決辦法。2.累加法在數(shù)列求和中的應(yīng)用和優(yōu)勢有哪些?3.舉例說明累加法在實際問題中的應(yīng)用,并分析其作用。4.與其他求數(shù)列通項公式的方法相比,累加法有什么特點和局限性？答案一、單項選擇題1.C2.C3.A4.D5.A6.B7.C8.A9.C10.C二、多項選擇題1.ABC2.ABCD3.ABCD4.BCD5.AB6.ABCD7.ABCD8.BCD9.ABCD10.ABD三、判斷題1.√2.×3.√4.√5.×6.√7.√8.√9.√10.√四、簡答題1.累加法適用于數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}-a_n=f(n)\)的形式,其中\(zhòng)(f(n)\)的前\(n\)項和可求,這樣通過依次累加\(a_2-a_1\),\(a_3-a_2\),\(\cdots\),\(a_n-a_{n-1}\)可求\(a_n\)。2.\(a_n-a_{n-1}=2(n-1)\),\(a_{n-1}-a_{n-2}=2(n-2)\),\(\cdots\),\(a_2-a_1=2\times1\)。累加得\(a_n-a_1=2\times[1+2+\cdots+(n-1)]=n(n-1)\),又\(a_1=1\),所以\(a_n=n^2-n+1\)。3.\(a_n-a_{n-1}=3^{n-1}\),\(a_{n-1}-a_{n-2}=3^{n-2}\),\(\cdots\),\(a_2-a_1=3^1\)。累加得\(a_n-a_1=3^1+3^2+\cdots+3^{n-1}\),由等比數(shù)列求和公式可得\(a_n=\frac{3^n-1}{2}\)。4.\(f(n)\)常見形式有：常數(shù)函數(shù),如\(f(n)=c\)（\(c\)為常數(shù)）；一次函數(shù),如\(f(n)=an+b\)；指數(shù)函數(shù),如\(f(n)=a^n\)；分式函數(shù),如\(f(n)=\frac{1}{n(n+1)}\)等。五、討論題1.困難:求和不易計算。解決辦法:嘗試對\(f(n)\)變形、利用已知求和公式、轉(zhuǎn)化為常見函數(shù)形式