高考數(shù)學(xué)二輪復(fù)習(xí)專題六數(shù)列微專題14數(shù)列中的創(chuàng)新性問題課件

微專題14數(shù)列中的創(chuàng)新性問題微專題14數(shù)列中的創(chuàng)新性問題

題型一新定義數(shù)列的創(chuàng)新型問題例1

(2017江蘇,19,16分)對于給定的正整數(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,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,an-2+an-1+an+1+an+2=4an,

①當(dāng)n≥4時,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),

③

an+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)新問題時,對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).解析(1)b1=1,b2=2,b3=3.(2)m為偶數(shù)時,{an}中有

項(xiàng)滿足題意,則bm=

;m為奇數(shù)時,{an}中有

項(xiàng)滿足題意,則bm=

,故bm=

m為偶數(shù)時,Sm=b1+b2+…+bm=

(1+2+…+m)-

·

=

;m為奇數(shù)時,Sm=Sm+1-bm+1=

-

=

.故Sm=

題型二新定義性質(zhì)型創(chuàng)新問題例2

(2018南師附中、天一、海門、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)公式及對應(yīng)的k的值;若不存在,請說明理由;(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,a1=a2-1,所以a2=2,故an+1=2an對任意的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,a1=a3-2,故a3=3,滿足a3=a2+a1,所以an+2=an+1+an對任意正整數(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