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微專(zhuān)題14數(shù)列中的創(chuàng)新性問(wèn)題,微專(zhuān)題14數(shù)列中的創(chuàng)新性問(wèn)題題型一新定義數(shù)列的創(chuàng)新型問(wèn)題,例1(2017江蘇,19,16分)對(duì)于給定的正整數(shù)k,若數(shù)列{an}滿足:an-k+an-k+1+…+an-1+an+1+…+an+k-1+an+k=2kan對(duì)任意正整數(shù)n(n>k)總成立,則稱數(shù)列{an}是“P(k)數(shù)列”.(1)證明:等差數(shù)列{an}是“P(3)數(shù)列”;(2)若數(shù)列{an}既是“P(2)數(shù)列”,又是“P(3)數(shù)列”,證明:{an}是等差數(shù)列.,證明(1)因?yàn)閧an}是等差數(shù)列,設(shè)其公差為d,則an=a1+(n-1)d,從而,當(dāng)n≥4時(shí),an-k+an+k=a1+(n-k-1)d+a1+(n+k-1)d=2a1+2(n-1)d=2an,k=1,2,3,所以an-3+an-2+an-1+an+1+an+2+an+3=6an,因此等差數(shù)列{an}是“P(3)數(shù)列”.(2)數(shù)列{an}既是“P(2)數(shù)列”,又是“P(3)數(shù)列”,因此,當(dāng)n≥3時(shí),an-2+an-1+an+1+an+2=4an,①當(dāng)n≥4時(shí),an-3+an-2+an-1+an+1+an+2+an+3=6an.②由①知,an-3+an-2=4an-1-(an+an+1),③an+2+an+3=4an+1-(an-1+an).④,將③④代入②,得an-1+an+1=2an,其中n≥4,所以a3,a4,a5,…是等差數(shù)列,設(shè)其公差為d.在①中,取n=4,則a2+a3+a5+a6=4a4,所以a2=a3-d,,在①中,取n=3,則a1+a2+a4+a5=4a3,所以a1=a3-2d,所以數(shù)列{an}是等差數(shù)列,,【方法歸納】解決數(shù)列新定義運(yùn)算型創(chuàng)新問(wèn)題時(shí),對(duì)新定義信息的提取和化歸轉(zhuǎn)化是解題的關(guān)鍵,也是解題的難點(diǎn).,1-1(2017江蘇泰州中學(xué)模擬)若數(shù)列{an}中不超過(guò)f(m)的項(xiàng)的個(gè)數(shù)恰為bm(m∈N*),則稱數(shù)列{bm}是數(shù)列{an}的生成數(shù)列,稱相應(yīng)的函數(shù)f(m)是數(shù)列{an}生成{bm}的控制函數(shù).(1)已知an=n2,且f(m)=m2,寫(xiě)出b1、b2、b3;(2)已知an=2n,且f(m)=m,求{bm}的前m項(xiàng)和Sm.,解析(1)b1=1,b2=2,b3=3.(2)m為偶數(shù)時(shí),{an}中有項(xiàng)滿足題意,則bm=;m為奇數(shù)時(shí),{an}中有項(xiàng)滿足題意,則bm=,故bm=m為偶數(shù)時(shí),Sm=b1+b2+…+bm=(1+2+…+m)-=;m為奇數(shù)時(shí),Sm=Sm+1-bm+1=-=.,故Sm=,題型二新定義性質(zhì)型創(chuàng)新問(wèn)題,例2(2018南師附中、天一、海門(mén)、淮陰四校聯(lián)考)設(shè)數(shù)列{an}的首項(xiàng)為1,前n項(xiàng)和為Sn,若對(duì)任意的n∈N*,均有Sn=an+k-k(k是常數(shù)且k∈N*)成立,則稱數(shù)列{an}為“P(k)數(shù)列”.(1)若數(shù)列{an}為“P(1)數(shù)列”,求數(shù)列{an}的通項(xiàng)公式;(2)是否存在數(shù)列{an}既是“P(k)數(shù)列”,也是“P(k+2)數(shù)列”?若存在,求出符合條件的數(shù)列{an}的通項(xiàng)公式及對(duì)應(yīng)的k的值;若不存在,請(qǐng)說(shuō)明理由;(3)若數(shù)列{an}為“P(2)數(shù)列”,a2=2,設(shè)Tn=+++…+,證明:Tn<3.,解析(1)數(shù)列{an}為“P(1)數(shù)列”,則Sn=an+1-1,,故Sn+1=an+2-1,兩式相減得an+2=2an+1,,又n=1時(shí),a1=a2-1,所以a2=2,故an+1=2an對(duì)任意的n∈N*恒成立,即=2(常數(shù)),故數(shù)列{an}為等比數(shù)列,其通項(xiàng)公式為an=2n-1,n∈N*.(2)假設(shè)存在這樣的數(shù)列{an},則有Sn=an+k-k,故有Sn+1=an+k+1-k,兩式相減得an+1=an+k+1-an+k,故有an+3=an+k+3-an+k+2,同理由{an}是“P(k+2)數(shù)列”可得:an+1=an+k+3-an+k+2,所以an+1=an+3對(duì)任意n∈N*恒成立,所以Sn=an+k-k=an+k+2-k=Sn+2,即Sn=Sn+2,又Sn+2=an+k+2-k-2=Sn-2,即Sn+2=Sn-2,,兩者矛盾,故不存在這樣的數(shù)列{an}既是“P(k)數(shù)列”,也是“P(k+2)數(shù)列”.,(3)證明:因?yàn)閿?shù)列{an}為“P(2)數(shù)列”,所以Sn=an+2-2,,所以Sn+1=an+3-2,,故有,an+1=an+3-an+2,又n=1時(shí),a1=a3-2,故a3=3,滿足a3=a2+a1,所以an+2=an+1+an對(duì)任意正整數(shù)n恒成立,數(shù)列的前幾項(xiàng)為:1,2,3,5,8,故Tn=+++…+=+++++…+,所以Tn=+++…++,兩式相減得:Tn=++++…+-=++++…+-=+Tn-2-,顯然Tn-20,故Tn<+Tn,即Tn<3.,【方法歸納】解決數(shù)列新定義性質(zhì)型創(chuàng)新問(wèn)題時(shí),需要細(xì)細(xì)品味新定義的概念、法則,對(duì)新定義的信息進(jìn)行加工,探求解決方法,有時(shí)可以尋找相近知識(shí)點(diǎn),明確它們的共同點(diǎn)和不同點(diǎn).,