板塊二數(shù)列微專題11數(shù)列中的最值、范圍問題高考定位近幾年高考試題中,與數(shù)列有關(guān)的最值范圍問題既有解答題,也有選擇、填空題,難度中檔或偏上.高考定位【

真題體驗(yàn)

】1.(2024·上海卷)等比數(shù)列{an}的首項(xiàng)a1>0,公比q>1,記In={x-y|x,y∈[a1,a2]∪

[an,an+1]},若對任意正整數(shù)n,In是閉區(qū)間,則q的取值范圍是

.

[2,+∞)顯然等比數(shù)列{an}遞增,不妨設(shè)x≥y,若x,y∈[a1,a2],則x-y∈[0,a2-a1],若x,y∈[an,an+1],則x-y∈[0,an+1-an],若x∈[an,an+1],y∈[a1,a2],則x-y∈[an-a2,an+1-a1],∵對任意正整數(shù)n,In都是閉區(qū)間,∴an-a2≤an+1-an,如圖,又a1>0,∴qn-2qn-1+q≥0,即qn-2(q-2)+1≥0,對任意正整數(shù)n,上式都成立,則必有q≥2.

(2)設(shè)數(shù)列{bn}滿足3bn+(n-4)an=0(n∈N*),記{bn}的前n項(xiàng)和為Tn.若Tn≤λbn對任意n∈N*恒成立,求實(shí)數(shù)λ的取值范圍.

精準(zhǔn)強(qiáng)化練熱點(diǎn)一求數(shù)列和式的最值、范圍熱點(diǎn)二求n的最值或范圍熱點(diǎn)三求數(shù)列不等式中參數(shù)的取值范圍熱點(diǎn)突破

熱點(diǎn)一求數(shù)列和式的最值、范圍例1

(2)設(shè)bn=logaan(a>0且a≠1),求數(shù)列{bn}的前n項(xiàng)和Tn的最值.

規(guī)律方法(2025·連云港調(diào)研)記數(shù)列{an}的前n項(xiàng)和為Sn,對任意n∈N*,有Sn=n(an+n-1).(1)證明:{an}是等差數(shù)列;訓(xùn)練1因?yàn)镾n=nan+n(n-1),①所以當(dāng)n≥2時(shí),Sn-1=(n-1)an-1+(n-1)(n-2),②

①-②可得an=nan-(n-1)an-1+2n-2,得(1-n)an=-(n-1)an-1+2(n-1),得an-an-1=-2,故{an}為等差數(shù)列.(2)若當(dāng)且僅當(dāng)n=7時(shí),Sn取得最大值,求a1的取值范圍.若當(dāng)且僅當(dāng)n=7時(shí),Sn取得最大值,故a1的取值范圍為(12,14).已知數(shù)列{an}是遞增的等比數(shù)列.設(shè)其公比為q,前n項(xiàng)和為Sn,且滿足a1+a5=34,8是a2與a4的等比中項(xiàng).(1)求數(shù)列{an}的通項(xiàng)公式;熱點(diǎn)二求n的最值或范圍例2因?yàn)?是a2與a4的等比中項(xiàng),所以a2a4=82=64,因?yàn)閿?shù)列{an}是遞增的等比數(shù)列,所以an=a1qn-1=2×2n-1=2n.(2)若bn=n·an,Tn是{bn}的前n項(xiàng)和,求使Tn-n·2n+1>-100成立的最大正整數(shù)n的值.

求n的值或最值,一般涉及數(shù)列的項(xiàng)或和的最值與范圍,通?；瘹w為解關(guān)于n的不等式,或根據(jù)數(shù)列的單調(diào)性求解.規(guī)律方法(2025·丹東段測)記Sn為等差數(shù)列{an}的前n項(xiàng)和,4Sn=anan+1+1,an≠0,n∈N*.(1)求{an}的通項(xiàng)公式;訓(xùn)練2因?yàn)閧an}為等差數(shù)列,且4Sn=anan+1+1,an≠0,n∈N*,所以當(dāng)n≥2時(shí),有4Sn-1=an-1an+1,兩式相減,得4an=an(an+1-an-1)=2dan(d為等差數(shù)列{an}的公差),解得d=2.當(dāng)n=1時(shí),有4S1=a1a2+1,即4a1=a1a2+1,4a1=a1(a1+2)+1,解得a1=1.所以an=a1+(n-1)d=1+2(n-1)=2n-1.

(2025·重慶質(zhì)檢)已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足2Sn+1-Sn=2,a1=1.(1)求{an}的通項(xiàng)公式;熱點(diǎn)三求數(shù)列不等式中參數(shù)的取值范圍例3

求數(shù)列不等式中參數(shù)的取值范圍問題要看清楚是恒成立,還是有解問題,若f(n)≥M恒成立,則f(n)min≥M;若f(n)≥M有解,則f(n)max≥M.易錯(cuò)提醒

訓(xùn)練3

(2)若對任意的n∈N*,都有an≤a6成立,求k的取值范圍.

【精準(zhǔn)強(qiáng)化練】√1.(2025·哈爾濱模擬)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若a4+a7<0,S9>0,則Sn的最大值為 A.S4 B.S5 C.S6 D.S7由S9=9a5>0,得a5>0,又a5+a6=a4+a7<0,則a6<0,所以公差d=a6-a5<0,故當(dāng)n≤5時(shí),an>0,當(dāng)n≥6時(shí),an<0,所以當(dāng)n=5時(shí),Sn最大.√

√

設(shè)數(shù)列{an}的公比q,因?yàn)?<m<1,an>0,所以q>0,S3=m+mq+mq2=7m,解得q=2,

√

當(dāng)n≤4時(shí),數(shù)列{an}單調(diào)遞減,且an<1;當(dāng)n≥5時(shí),數(shù)列{an}單調(diào)遞減,且an>1,故an的最大值為a5=3,最小值為a4=-1,∴an的最大值與最小值之和為2,故選C.√

√6.(2025·Z20名校聯(lián)盟二聯(lián))定義在(0,+∞)的增函數(shù)f(x)滿足:f(x)+f(y)=f(xy)-1,且f(2)=0,f(an)=n-1.已知數(shù)列{an}的前n項(xiàng)和為Sn,則使得Sn<2025成立的n的最大值是 A.8 B.9 C.10 D.11法一

∵f(x)+f(y)=f(xy)-1,可令f(x)=logax-1,又f(2)=0,則loga2-1=0,∴a=2,∴f(x)=log2x-1.∵f(an)=log2an-1=n-1,∴an=2n,

√

√

若an=2n-1,則an+2+an-2an+1=2n+3+2n-1-2(2n+1)=0,即an+2+an=2an+1,不滿足條件,不是“M-數(shù)列”;若an=-3n,則an+2+an-2an+1=-(3n+2+3n-2×3n+1)=-4×3n<0,即an+2+an<2an+1,不滿足條件,不是“M-數(shù)列”;若an=n×2n,則an+2+an-2an+1=(n+2)×2n+2+n×2n-2(n+1)×2n+1=(n+4)×2n>0,

√

√√

√

√√因?yàn)閧3n-1an}的前n項(xiàng)和為n·3n,所以有30a1+31a2+32a3+…+3n-1an=n·3n,顯然a1=3,顯然當(dāng)n≥2,n∈N*時(shí),有30a1+31a2+32a3+…+3n-2an-1=(n-1)·3n-1,

√

√√

5

f'(x)=4an+1x3-3anx2-an+2,∴f'(1)=4an+1-3an-an+2=0,即an+2-an+1=3(an+1-an),∴數(shù)列{an+1-an}是首項(xiàng)為2,公比為3的等比數(shù)列,1012

又因?yàn)閍1=2,a2=6,所以a2-