Question

已知數(shù)列{a_n}:a₁=1,a₂=2,a₃=r,a_n+3=a_n+2(n是正整數(shù)),與數(shù)列{b_n}:b₁=1,b₂=0,b₃=-1,b₄=0,b_n+4=b_n(n是正整數(shù)).

記T_n=b₁a₁+b₂a₂+b₃a₃+…+b_na_n.

(1)若a₁+a₂+a₃+…+a₁₂=64,求r的值;

(2)求證:當(dāng)n是正整數(shù)時,T_12n=-4n;

(3)已知r＞0,且存在正整數(shù)m,使得在T_12m+1,T_12m+2,…,T_12m+12中有4項為100,求r的值,并指出哪4項為100.

Answer 1

(1)解:a₁+a₂+a₃…+a₁₂

=1+2+r+3+4+(r+2)+5+6+(r+4)+7+8+(r+6)

=48+4r.                                                                     

∵48+4r=64,∴r=4.                                                           

(2)證明:用數(shù)學(xué)歸納法證明:當(dāng)n∈Z⁺時,T_12n=-4n.

①當(dāng)n=1時,T₁₂=a₁-a₃+a₅-a₇+a₉-a₁₁=-4,等式成立.                                  

②假設(shè)n=k時等式成立,即T_12k=-4k,

那么當(dāng)n=k+1時,

T_12(k+1)=T_12k+a_12k+1-a_12k+3+a_12k+5-a_12k+7+a_12k+9-a_12k+11                                 

=-4k+(8k+1)-(8k+r)+(8k+4)-(8k+5)+(8k+r+4)-(8k+8)

=-4k-4=-4(k+1),等式也成立.

根據(jù)①和②可以斷定:當(dāng)n∈Z⁺時,T_12n=-4n.                                      

(3)解:T_12m=-4m(m≥1).

當(dāng)n=12m+1,12m+2時,T_n=4m+1;

當(dāng)n=12m+3,12m+4時,T_n=-4m+1-r;

當(dāng)n=12m+5,12m+6時,T_n=4m+5-r;

當(dāng)n=12m+7,12m+8時,T_n=-4m-r;

當(dāng)n=12m+9,12m+10時,T_n=4m+4;

當(dāng)n=12m+11,12m+12時,T_n=-4m-4.                                            

∵4m+1是奇數(shù),-4m+1-r,-4m-r,-4m-4均為負(fù)數(shù),

∴這些項均不可能取到100.                                                   

∴4m+5-r=4m+4=100,解得m=24,r=1.

此時T₂₉₃,T₂₉₄,T₂₉₇,T₂₉₈為100.