素能培優(yōu)(九)子數(shù)列問(wèn)題高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)GAOKAOZONGFUXIYOUHUASHEJI2026子數(shù)列問(wèn)題主要包括數(shù)列中的公共項(xiàng)、并項(xiàng)和增減項(xiàng)問(wèn)題.數(shù)列中的增減項(xiàng),是指由已知的數(shù)列通過(guò)插入項(xiàng)、剔除項(xiàng)或合并項(xiàng)等得到新的數(shù)列,再研究新數(shù)列的求和等問(wèn)題,是高考命題的熱點(diǎn).解決此類問(wèn)題的關(guān)鍵,是弄清楚新生成的數(shù)列與已知數(shù)列的關(guān)系,確定其特征,并根據(jù)題意和要求解題.題型一兩數(shù)列的公共項(xiàng)例1(1)(2020·新高考Ⅰ,14)將數(shù)列{2n-1}與{3n-2}的公共項(xiàng)從小到大排列得到數(shù)列{an},則{an}的前n項(xiàng)和為

.3n2-2n

(2)(2024·河南濮陽(yáng)高三質(zhì)檢)由數(shù)列{an}和{bn}的公共項(xiàng)組成的數(shù)列記為{cn},已知an=3n-2,bn=2n,若{cn}為遞增數(shù)列,且c3=bx=ay,則x+y=

.28

[對(duì)點(diǎn)訓(xùn)練1](2024·山東濟(jì)寧模擬)我國(guó)古代數(shù)學(xué)名著《孫子算經(jīng)》中有一道數(shù)學(xué)問(wèn)題:“今有物不知其數(shù),三三數(shù)之剩二,五五數(shù)之剩二,七七數(shù)之剩二,問(wèn)物幾何?”若所有被3除余2的自然數(shù)從小到大組成數(shù)列{an},所有被5除余2的自然數(shù)從小到大組成數(shù)列{bn},把{an}和{bn}的公共項(xiàng)從小到大排列得到數(shù)列{cn},則下列說(shuō)法正確的是(

)A.a3+b5=c3

B.b28=c10C.a5b2>c8

D.c9-b9=a26B解析

根據(jù)題意,數(shù)列{an}是首項(xiàng)為2,公差為3的等差數(shù)列,an=2+3(n-1)=3n-1;數(shù)列{bn}是首項(xiàng)為2,公差為5的等差數(shù)列,bn=2+5(n-1)=5n-3.數(shù)列{an}與{bn}的公共項(xiàng)從小到大排列得到數(shù)列{cn},故數(shù)列{cn}是首項(xiàng)為2,公差為15的等差數(shù)列,cn=2+15(n-1)=15n-13.a3+b5=(3×3-1)+(5×5-3)=30,c3=15×3-13=32,a3+b5≠c3,故A錯(cuò)誤;b28=5×28-3=137,c10=15×10-13=137,b28=c10,故B正確;a5=3×5-1=14,b2=5×2-3=7,c8=15×8-13=107,a5b2=14×7=98<107=c8,故C錯(cuò)誤;c9=15×9-13=122,b9=5×9-3=42,a26=3×26-1=77,c9-b9=122-42=80≠77=a26,故D錯(cuò)誤.故選B.題型二數(shù)列中的增減項(xiàng)問(wèn)題

[對(duì)點(diǎn)訓(xùn)練2](2024·廣東佛山模擬)若數(shù)列{an}滿足a1=1,an+1=3an+1,若bn=2an+1(n∈N*),抽去數(shù)列{bn}的第3項(xiàng)、第6項(xiàng)、第9項(xiàng)…余下的項(xiàng)的順序不變,構(gòu)成一個(gè)新數(shù)列{cn},則數(shù)列{cn}的前