培優(yōu)拓展(八)數(shù)列中的增減項(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.角度一數(shù)列中的減項(xiàng)問(wèn)題例1已知數(shù)列{an}是等差數(shù)列,前n項(xiàng)和為Sn;數(shù)列{bn}是各項(xiàng)均為正數(shù)的等比數(shù)列,前n項(xiàng)和為Tn;且a2=b2=4,a8=b4=16.(1)分別求數(shù)列{an},{bn}的通項(xiàng)公式和前n項(xiàng)和Sn,Tn;(2)若將數(shù)列{an}中出現(xiàn)的數(shù)列{bn}的項(xiàng)剔除后,剩余的項(xiàng)從小到大排列得到數(shù)列{cn},記數(shù)列{cn}的前n項(xiàng)和為Kn,求K2021.(2)將數(shù)列{an}中出現(xiàn)的數(shù)列{bn}的項(xiàng)剔除后,剩余的項(xiàng)從小到大排列得到數(shù)列{cn},∵b11=211=2

048<a2

021=4

042,b12=212=4

096>a2

021=4

042,b12=212=4

096>a2

032=4

064,∴數(shù)列{cn}的前2

021項(xiàng)和需要從數(shù)列{an}的前2

032項(xiàng)和中剔除{bn}的前11項(xiàng),∴K2

021=S2

032-T11=2

0322+2

032-212+2=4

126

962.角度二數(shù)列中的并項(xiàng)問(wèn)題例2設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,a2=7,且對(duì)任意的自然數(shù)n,恒有an+3=(1)求數(shù)列{an}的通項(xiàng)公式;(2)若集合A={x|x=an,n∈N*},B={x|x=3n,n∈N*},將集合A∪B中的所有元素按從小到大的順序排列構(gòu)成數(shù)列{bn},記數(shù)列{bn}的前n項(xiàng)和為Tn,求T102的值.可得n=1時(shí),a1=2a1-3,解得a1=3;n=2時(shí),2a2=2S2-6=2(a1+a2)-6,解得a1=3;n=3時(shí),3a3=2S3-9=2(a1+a2+a3)-9,解得a3=11.當(dāng)n≥2時(shí),由an+3=,得nan=2Sn-3n,得(n-1)an-1=2Sn-1-3(n-1),兩式相減可得(n-2)an=(n-1)an-1-3,當(dāng)n≥3時(shí),由上式得(n-3)an-1=(n-2)·an-2-3,上面兩式相減可得an+an-2=2an-1,n≥3,且a1+a3=2a2,所以數(shù)列{an}是首項(xiàng)為3,公差為4的等差數(shù)列,所以an=3+4(n-1)=4n-1.(2)因?yàn)榧螦={x|x=4n-1,n∈N*},B={x|x=3n,n∈N*},所以集合A∪B中的所有元素的最小值為3,因?yàn)?5=243<a100=399,且3,32,33,34,35中,3,33,35三個(gè)元素是{bn}前102項(xiàng)中的元素,也是A∩B中的元素,所以T102=a1+a2+a3+…+a100+9+81=×100×(3+400-1)+90=20

190.角度三數(shù)列中的插項(xiàng)問(wèn)題例3(2024江蘇南京模擬)已知各項(xiàng)均為正數(shù)的數(shù)列{an}中,a1=1且滿足

=2an+2an+1,數(shù)列{bn}的前n項(xiàng)和為Sn,滿足2Sn+1=3bn.(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)若在bk與bk+1之間依次插入數(shù)列{an}中的k項(xiàng)構(gòu)成新數(shù)列{cn}:b1,a1,b2,a2,a3,b3,a4,a5,a6,b4,…,求數(shù)列{cn}中前50項(xiàng)的和T50.針對(duì)訓(xùn)練1.(2024廣東梅州一模)設(shè){an}是等差數(shù)列,{bn}是等比數(shù)列.已知a1=b1=4,b2=a2+1,b3=2a3-4.(1)求{an}和{bn}的通項(xiàng)公式;(2)數(shù)列{an}和{bn}的項(xiàng)從小到大依次排列(相等項(xiàng)計(jì)兩項(xiàng))得到新數(shù)列{cn},求{cn}的前50項(xiàng)的和.解

(1)設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q,所以an=3n+1,bn=2n+1.(2)由an=3n+1,bn=2n+1,可得{cn}的前50項(xiàng)中有3個(gè)相同的項(xiàng),為4,16,64.所以當(dāng)數(shù)列{cn}的前50項(xiàng)中含有數(shù)列{bn}的前5項(xiàng)時(shí),令3n+1=25+1=64,得n=21,則{cn}的第26項(xiàng)為64,當(dāng)數(shù)列{cn}的前50項(xiàng)中含有數(shù)列{bn}的前6項(xiàng)時(shí),令3n+1