2025優(yōu)化設(shè)計(jì)一輪課時(shí)規(guī)范練48　數(shù)列中的增項(xiàng)、減項(xiàng)問(wèn)題

課時(shí)規(guī)范練48數(shù)列中的增項(xiàng)、減項(xiàng)問(wèn)題1.(2024·山東泰安模擬)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,數(shù)列{bn}為等比數(shù)列,滿足a1=b2=2,S5=30,b4+2是b3與b5的等差中項(xiàng).(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)從數(shù)列{an}中去掉數(shù)列{bn}的項(xiàng)后余下的項(xiàng)按原來(lái)的順序組成數(shù)列{cn},設(shè)數(shù)列{cn}的前n項(xiàng)和為T(mén)n,求T60.2.(2024·廣東佛山模擬)已知數(shù)列{an}滿足a13+a232+…+a(1)求{an}的通項(xiàng)公式;(2)在{an}相鄰兩項(xiàng)的中間插入這兩項(xiàng)的等差中項(xiàng),求所得新數(shù)列{bn}的前2n項(xiàng)和T2n.3.(2024·天津耀華中學(xué)模擬節(jié)選)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn=(n-1)2n+1+2,n∈N*.(1)求{an}的通項(xiàng)公式;(2)若bn=ann,抽去數(shù)列{bn}中的第1項(xiàng),第4項(xiàng),第7項(xiàng),…,第3n-2項(xiàng),…,余下的項(xiàng)順序不變,組成一個(gè)新數(shù)列{cn},若{cn}的前n項(xiàng)和為T(mén)n,求證:當(dāng)n為奇數(shù)時(shí),1254.(2024·遼寧丹東模擬)Sn為數(shù)列{an}的前n項(xiàng)和,已知Sn=3-an-n2.(1)證明:an=3-2n;(2)保持?jǐn)?shù)列{an}中各項(xiàng)先后順序不變,在ak與ak+1之間插入數(shù)列{(n+1)2n}的前k項(xiàng),使它們和原數(shù)列的項(xiàng)構(gòu)成一個(gè)新的數(shù)列a1,2×21,a2,2×21,3×22,a3,2×21,3×22,4×23,a4,…,求這個(gè)新數(shù)列的前50項(xiàng)和.

課時(shí)規(guī)范練48數(shù)列中的增項(xiàng)、減項(xiàng)問(wèn)題1.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.∵a1=2,S5=10+5×42d=30,∴an=2+2(n-1)=2n.∵b4+2是b3與b5的等差中項(xiàng),∴2(b4+2)=b3+b5.又b2=2,∴2(2q2+2)=2q+2q3,解得q=2,∴bn=2n-1.(2)∵a60=120,∴數(shù)列{an}的前60項(xiàng)中與數(shù)列{bn}的公共項(xiàng)共有6項(xiàng),且最大公共項(xiàng)為b7=26=64.又a66=132,b8=27=128,∴T60=S67-(2+22+…+27)=134+67×662×2-2×(1-27)2.解(1)因?yàn)閍13+a232+…+an所以當(dāng)n≥2時(shí),a13+a232①-②得an3n=n3?n-13=1又當(dāng)n=1時(shí),a13=13,所以a1=1也滿足上式,故{an}的通項(xiàng)公式為an(2)設(shè)數(shù)列{cn}滿足cn=an+an+12=記{an}的前n項(xiàng)和為Sn,{cn}的前n項(xiàng)和為Rn,則T2n=Rn+Sn.由等比數(shù)列的求和公式得Sn=1-3n1-3=12(3n-1),Rn=所以T2n=Rn+Sn=32(3n-1).即新數(shù)列{bn}的前2n項(xiàng)和T2n=32(3n-3.(1)解由Sn=(n-1)2n+1+2,得Sn-1=(n-2)2n+2(n≥2),兩式相減得an=n·2n,n≥2.當(dāng)n=1時(shí),a1=2滿足上式,所以an=n·2n.(2)證明由(1)知,bn=ann=2n,所以數(shù)列{cn}為22,23,25,26,28,29,它的奇數(shù)項(xiàng)組成以4為首項(xiàng),8為公比的等比數(shù)列;偶數(shù)項(xiàng)組成以8為首項(xiàng),8為公比的等比數(shù)列.∴當(dāng)n為奇數(shù),即n=2k-1(k∈N*)時(shí),Tn=(c1+c3+…+c2k-1)+(c2+c4+…+c2k-2)=4(1-8kTn+1=Tn+cn+1=57×8k-127+23k=127∵5×8k-12≥28,∴1254.(1)證明由S1=3-a1-1,得a1=1.由Sn=3-an-n2,可知Sn+1=3-an+1-(n+1)2.兩式相減得2an+1=an-2n-1,所以an+1+2n-1=12(an+2n-3)又a1-1=0,所以an+2n-3=0,因此an=3-2n.(2)解設(shè)數(shù)列{(n+1)2n}的前n項(xiàng)和為T(mén)n,則Tn=2×21+3×22+…+(n+1)2n,2Tn=2×22+3×23+4×24+…+(n+1)2n+1.兩式相減得Tn=(n+1)2n+1-4-(22+23+…+2n)=n2n+1.設(shè)ak+1是新數(shù)列的第N項(xiàng),則N=1+2+…+(k+1)=(當(dāng)k=8時(shí),N=45<50,當(dāng)k=9時(shí),N=55>50.故這個(gè)新數(shù)列的前50項(xiàng)和中包含{an}的前9項(xiàng)、Tk(k=1,2,3,…,8)以及數(shù)列{(n+1)2n}的前5項(xiàng).由(1)知Sn=n(2-n),所以這個(gè)新數(shù)列的前50項(xiàng)和為S9+(1×22+2×23