用函數(shù)求數(shù)列通項與和

用函數(shù)[x]求數(shù)列通項與和(省開江普安中學(xué)函數(shù)f(x)=[x](x∈R)表示不超過實 S12mm-

)=12+22+32+?+(m-1)x的最大整數(shù)部分,稱函數(shù)[x]叫做函數(shù).利用它求某些較難的數(shù)列通項與n項之和卻有.本文以實例來說明用高斯函數(shù)[x]求數(shù)列通項與和的方法,供參考.

=1m(m-1)(2m-1)6an=m,則數(shù)列an}n212例 設(shè)數(shù)列{an}的各項

Sn=S1m(m-1) n-2m(m-1)1,2,2,3,3,3,?,n,n,?,n,anSn

=1m(m-1)(2m-1)+6-1m2(m-1):,它是各段項數(shù)越來越多(呈等差數(shù)列的分段遞增數(shù)列.

=mn-1(

m+1)(m-1)設(shè)數(shù)列{an}的通項an=m(m為段數(shù)列各段的順序數(shù))由等差數(shù)列求和公,得前(m-1段項數(shù)之和為

其中m=1 例 設(shè)數(shù)列{an}的各項1+2+3+

+(m-1)=1m(m-1)2

1,1,2,2,2,2,2,?,n,n,?,n,(3n-11m(m-1)+12

1m(m+ ≤2≤

anSn解:由等差數(shù)列求和,得數(shù)列{an}前(m-1)段項數(shù)的和為時,an=m.解關(guān)于m的不等式(1), 2+15+8+?+(3m--1 8n+1 1 8n-7

(m-1)(3m-2)<1相同的項有m(m=1,2,?)個<1∵01 8n-7--1 8n+

(m-1)(3m-2)+1

1m(3m+ 1 8n-

2∴m 時,an=m.解關(guān)于m的不等式(2),1 -1 24n+1 5 24n-23m 故a=1 8n-7 相同的項有(3m-1)(m=1,2,?)個 5 24n- -1 24n+n于是,數(shù)列{a}前(m-1)段之和 <1n 本文收稿日期

∴m為1(5 24n-23)的整數(shù)部分記作m=5 ∴m為1 n-1的整數(shù)部分,記m=[1 n-1]故an=5

an3[1

n-1]-,數(shù)列an}n-=3(12+22+?+n2)-(1+2+?+

,數(shù)列an}n-=6(12+?+n2)-5(1+2+?+n)+ =3×6n(n+1)(2n+1)=n2(n+1)

2n(n+

=1n(4n2+n-1)2n改為m-1),得前m-1S1 ) )=m(m-1)2

nm-1得前m-112m-13m-

2 (m-1)(4m2-7m+2)∵an=m

(m- ∵an=3m-1nn-1(m-1)(3m-2)

2Sn=S1(m-1)(3m-2)2

Sn=S(m-1)2+[n-(m-1)2](3m- =1(m-1)(4m2-7m+2)+3mn-=m(m-1)2+mn1

2m(m-1)(3m-

-(3m-1)(m-1)=mn m2(m-1) =3mn-n5

1m(m-1)(2m-1)2其中m

其中m=[1 ]注:,只有深刻理解數(shù)列{an}n用以確定項數(shù)(或段數(shù)和確定,n既確定求和中各項(或段)的元素也決定著它的項數(shù)(或段數(shù))的雙重作用,才能正確解答此類問題.3設(shè)數(shù)列an}2,5,5,5,?,(3n-1),?,(3n-1),(2n-1anSn解:由等差數(shù)列求和,得數(shù)列{an}的前(m-1)段項數(shù)之和為1+3+5+?+(2m-3)=(m-1)2(m-1)2+1≤n an3m-1.m的不等式(3),

例 設(shè)數(shù)列{an}的各項3,3,7,7,?,(4n-1),(4n-1),anSn:,即每一段(重復(fù)出現(xiàn)的幾個項)的項數(shù)都相等的遞增數(shù)列.由等差數(shù)列求和,得數(shù)列{an}前m-12+2+?+2=2(m-1)m-12(m-1)+1≤n≤2 an4m-1m的不等式(4),≤n n1,2個.≤ ∵0<n+1-n<1 n n-1nn相同的項有(2m-1)(m12?個n

n+

n-1 <1 m 2∴an=4n+2

- nm-1得前m-11,數(shù)列an}n1Sn2[3711+?+(4n-1)=2n[3+(4n-1)]=2n(2n+1)

Sk(m-1)=2km(m-1)∵an=m·

=Sk(m-

+[n-k(m-1)]nm-1得前m-1S2(m-1)=2(m-1)(2m-1)∵an=4m-1Sn=S2(m-1)+[n-2(m-1)](4m-=2(m-1)(2m-1)+4mn--(4m-1)2(m-=4mn-n-4m(m-1)

=km(m-1)+mn-km(m-1)=mn-1km(m-1)2其中m n+k-1k由例5可以,有如下題命題由函數(shù)f(n) n+k-r所kmn+121,1,?,1,2,2,?,2,?,n,n,?,n, anSn

定的數(shù)列f(n},有以下形式第1 第2 第n0,0,?,0,1,1,?,1,?,n,n,?,n,(r-1) :1≤n<r,k≥r≥10<1+1-r≤n+k-r<1(m-1)段項數(shù)的和 k+k+?+k=km-1 n12,?r-1,m-1

f(n) n+k-k

01組k(m-1)+1≤n an=mm的不等式(5),n≤m≤n+k-1,mm 1,2,?)個∵0<n+k-1-n<1

r≤n<k+r, n+k- <2n=rr1,?k+r-1f(n)12組 ∴m為n+k-1,km n+k-1k

若在定理中允許r>k,則第1組有r-(k+1),后面各組有k,且后一組的數(shù)比前一組的數(shù)大1.利用命題對k≥rk故an n+k-1 有k,數(shù)列an}n++

①r1k2得數(shù)列1,1,2,2,3,3,4,4?.

n+1 =k(1+2+?+n)=1kn(n+1) an n+1,Sn=mn-m(m-1)2其中m n+12②r=1,k=3,得數(shù)列111222333?.

n+2

n,本題可視為設(shè)數(shù)列an}1234an=an1n12?),Sn,可用如下的解法an= n+2

,

=mn

3m(m-1)2

im∈Nki{12,34}m,n=4m+in+其中m

:Sn=10+14+18+?+(4m+:

n+k-k

+(m+1)i+1,1,?,1,2,2,?,2,3,3,?,3,1 1

=2m(m+4)+(m+1)i+=2·n-in-i+4+n-i+1i+a n+k-1,S=mn n+k-

km(m-1)2

n2+16n+i2-8i+8 其中m n+k-

n2+16n-

i2+8i-8k-

n2+16 i2+8i-80,1,1,?,1,2,2,?,2,3,3,?,3, i2+8i-8k- an n+k-2k

0,1,3)

<1(i=1,2,3;kSn=mn-m-1km(m-1)2

i2+8i-80 <1(i=4,k=6)k其中m n+k-k

運用函數(shù)[x]求數(shù)列的通項前n項之和的問題,是一種新穎的求 Sn

n2+16n-82

,

n時

(n8方法.這種方法非常巧妙,且推導(dǎo)自然,結(jié) n+168

,4|n時例 設(shè)數(shù)列{an}的各項 ∴S1921

81,2,3,4,2,3,4,5,3,4,5,6, 3720

,

S1

之值

=465:1,2,3,4,5,6,7,8,9,10,11,12?n,0000333366

1,我們巧設(shè)了待定常數(shù)i,m之值是設(shè)而不求. n-1,對應(yīng)項相減,得 m和i來表示n,并同時用m和ik(非零4123423453456?

,m=n-i,4n-n-

-gik,0≤g(i4an=n-3n-1480-

k)1從而利用[x求出

=[f(n)](na80=80- =

注2:用兩個函數(shù)[x]相加(減)的例 設(shè)數(shù)列{an}的各項

k3Sn=mn-m-3mm-1230,1,1,?,12,2,?,233,?,3, m=23

=667 求an及Sn,并當k=3時,計算a2002與S2 ∴S2002=667×2002-之值

-3×667×(667-2m21+k+k+?+k=k(m-1)+m-1k(m-1)+2≤n≤km+ an=mm的不等式(6),

=1335334-667-666333=668例 1,1,1,2,3,3,3,4,5,5,5,6,7,7,7,8,anSn:111122223333444 n+34n-1

n+k-( ,相同的項有mm( 0,0,0,1,1,1,1,2,2,2,2,3,3,312?個∵0<n+k-2-n-1<1

34?的通項為n4 ,1,1,1,2,3,33,4∴m為n+k-2的整數(shù)部分,記 5,5,5,6,7,7,7,8,?的通項k=.n+k-=.

n+4

n4an

n+k-2(k≥2)

故an n+4

n4kk3kan n+1,a20023

2 =

n,本題可視為設(shè)數(shù)列an}1112an=an2n12?Sn,數(shù)列an}nSn01·k2·k+?+n-1)

,有如下的解法im∈Nki=k[1+2

?+(n-1)]=kn(n-1)2

{123,4}m,n=4m+i則nm,mSk(m-1)+1=1km(m-1)2n+k-

Sn=5+13+21+?+(8m-+(2m+1)i+=m(4m+1)+(2m+1)i+∵an Sn=Sk(m-1)+1+[(n-1)-k(m-1)]

=n-i4·n-i+1+2·n-i+1i+ n2+n-i2+3i+4 =1k(m-1)m+mn-m-km(m-

n2+n+ i2-3i-4k+2=mn-m

km(m-1) i2-3i+2-4

2其中m n+k-2k

<1(i=1,2,3,4,00,0 EF,兩對角線的交點為P,過P⊥EFO求證:∠BOC=

8四、(a求所有自然數(shù)n(n≥2),a1a2?,an,{|ai-aj||1≤i<j(冷崗松肖振綱命題 =1,2,?,n(n- (許以超命題

=1,

=1(1+ n-

)2,n≥2.求最 4(bA={1,2,3,4,5,6B={7,8,9,?n4實數(shù)λ,使得對任意非負實數(shù)x1,x2,?,x2002,都 A中取三個數(shù)、B中取兩個數(shù)組成五個元素的集2

Aii1,2,?,20|A∩A|≤2,1≤i<jk=A

xk- 22,k 求n的最小值 (裘宗滬命題22kxk+?+x2002

k(k-1)+

kf

(李勝宏命題

,滿足f(n)f(n+1)=(f(n)+n-k)2n=-2,-3,-4,每人每場至多預(yù)訂一張門票

f(n的表達式

(陳永高命題3六、fxxx)=-2x+x3+x3)3x2 (x2+x3)+x2(x3