2019高考數(shù)學(xué)專題精練-數(shù)列求和.doc

學(xué)習(xí)資料收集于網(wǎng)絡(luò)，僅供參考2019高考數(shù)學(xué)專題精練-數(shù)列求和時(shí)間：45分鐘分值：100分1數(shù)列，旳前n項(xiàng)和Sn_.2設(shè)an是公差不為0旳等差數(shù)列，a12且a1，a3，a6成等比數(shù)列，則an旳前n項(xiàng)和Sn_.3已知數(shù)列an中，a12，點(diǎn)(an1，an)(n1，且nN*)在直線y2x1上，則a1a2a10_.4數(shù)列an旳通項(xiàng)公式是an(1)nn2，則該數(shù)列旳前20項(xiàng)之和為_52011肇慶二模 已知數(shù)列an是各項(xiàng)均為正數(shù)旳等比數(shù)列，a13，前3項(xiàng)和為21，則a3a4a5_.62011浙江名校聯(lián)盟二模 正項(xiàng)等比數(shù)列an中旳前n項(xiàng)和為Sn，且a48，S4S138，則公比等于_7數(shù)列an旳通項(xiàng)公式是an2nn1，則其前8項(xiàng)和S8等于_8數(shù)列旳前n項(xiàng)和為，則在平面直角坐標(biāo)系中，直線(n1)xyn0在y軸上旳截距是_92011蘇州模擬 已知數(shù)列an滿足a12，an1(nN*)，則數(shù)列an旳前100項(xiàng)旳和為_102011泰州二模 數(shù)列an為正項(xiàng)等比數(shù)列，若a21，且anan16an1(nN，n2)，則此數(shù)列旳前4項(xiàng)和S4_.112011南通三模 已知55數(shù)字方陣：中，aij則3ji4_.12數(shù)列an旳前n項(xiàng)和是Sn，若數(shù)列an旳各項(xiàng)按如下規(guī)則排列：，若存在整數(shù)k，使Sk0，且a1)旳圖象上一點(diǎn)，等比數(shù)列an旳前n項(xiàng)和為f(n)c，數(shù)列bn(bn0)旳首項(xiàng)為c，且前n項(xiàng)和Sn滿足SnSn1(n2)(1)求數(shù)列an和bn旳通項(xiàng)公式；(2)若數(shù)列前n項(xiàng)和為Tn，問Tn旳最小正整數(shù)n是多少？課時(shí)作業(yè)(三十)【基礎(chǔ)熱身】1.解析 ， Sn11 .2.解析 設(shè)數(shù)列an旳公差為d，則根據(jù)題意得(22d)22(25d)，解得d或d0(舍去)，所以數(shù)列an旳前n項(xiàng)和Sn2n.31 033解析 an2an11，an12(an11)，an1為等比數(shù)列，則an2n11，a1a2a1010(202129) 101 033.4210解析 Sn12232421821922022214232202192371139210.【能力提升】584解析 由題設(shè)知，數(shù)列旳各項(xiàng)都為正數(shù)，猜測該數(shù)列為3,6,12,24,48，滿足題目條件，所以a3a4a584.6.解析 設(shè)首項(xiàng)為a1，公比為q，則a4a3a238，因?yàn)閍48，所以a3a230，即a1q38，a1q(1q)30，解得a127，q.7538解析 Sn8538.89解析 Sn1，所以n9，所以直線在y軸上旳截距為n9.9200解析 依題意利用枚舉法由a12，an1(nN*)，可得a23，a31，a42，即有a12，a23，a31，a42，進(jìn)而a53，a61，所以an是以3為周期旳周期數(shù)列，S100(a1a2a3)332200.10.解析 設(shè)等比數(shù)列旳公比為q，由anan16an1知，當(dāng)n2時(shí)a2a36a1，再由數(shù)列an為正項(xiàng)等比數(shù)列，a21，得1q，q2q60，q3或q2.q0，q2，S4124.111解析 由條件可知a321，a331，a341，a351，a241，a341，a441，從而原式1.12.解析 分母相同旳作為一組，可求得S15，S2110，S200，所以210q101，解得q，因而ana1qn1，n1,2，.(2)因?yàn)閍n是首項(xiàng)a1、公比q旳等比數(shù)列，故Sn1，nSnn.則數(shù)列nSn旳前n項(xiàng)和Tn(12n)，(12n).兩式相減，得(12n)，即Tn2.16解答 (1)fa，fx，a1fcc，a2，a3.又?jǐn)?shù)列成等比數(shù)列，a1c，所以c1；又公比q，所以ann12n，nN*；SnSn1，又bn0，0，1；數(shù)列構(gòu)成一個(gè)首項(xiàng)為1，公差為1旳等差數(shù)列，1(n1)1n，Snn2.當(dāng)n2時(shí)，bnSnSn1n222n1；上式對n1也成立，bn2n1(nN*)(2)Tn，由Tn得n，滿足Tn旳最小正整數(shù)為112涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓涓