Question

已知数列{a_n}:a₁=1,a₂=2,a₃=r,a_n+3=a_n+2(n是正整数),与数列{b_n}:b₁=1,b₂=0,b₃=-1,b₄=0,b_n+4=b_n(n是正整数).

记T_n=b₁a₁+b₂a₂+b₃a₃+…+b_na_n.

(1)若a₁+a₂+a₃+…+a₁₂=64,求r的值;

(2)求证:当n是正整数时,T_12n=-4n;

(3)已知r＞0,且存在正整数m,使得在T_12m+1,T_12m+2,…,T_12m+12中有4项为100,求r的值,并指出哪4项为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)证明:用数学归纳法证明:当n∈Z⁺时,T_12n=-4n.

①当n=1时,T₁₂=a₁-a₃+a₅-a₇+a₉-a₁₁=-4,等式成立.                                  

②假设n=k时等式成立,即T_12k=-4k,

那么当n=k+1时,

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.

根据①和②可以断定:当n∈Z⁺时,T_12n=-4n.                                      

(3)解:T_12m=-4m(m≥1).

当n=12m+1,12m+2时,T_n=4m+1;

当n=12m+3,12m+4时,T_n=-4m+1-r;

当n=12m+5,12m+6时,T_n=4m+5-r;

当n=12m+7,12m+8时,T_n=-4m-r;

当n=12m+9,12m+10时,T_n=4m+4;

当n=12m+11,12m+12时,T_n=-4m-4.                                            

∵4m+1是奇数,-4m+1-r,-4m-r,-4m-4均为负数,

∴这些项均不可能取到100.                                                   

∴4m+5-r=4m+4=100,解得m=24,r=1.

此时T₂₉₃,T₂₉₄,T₂₉₇,T₂₉₈为100.