题目

已知数列{an}:a1=1,a2=2,a3=r,an+3=an+2(n是正整数),与数列{bn}:b1=1,b2=0,b3=-1,b4=0,bn+4=bn(n是正整数).记Tn=b1a1+b2a2+b3a3+…+bnan.(1)若a1+a2+a3+…+a12=64,求r的值;(2)求证:当n是正整数时,T12n=-4n;(3)已知r＞0,且存在正整数m,使得在T12m+1,T12m+2,…,T12m+12中有4项为100,求r的值,并指出哪4项为100. 答案：(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,T12n=-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T12k=-4k,那么当n=k+1时,T12(k+1)=T12k+a12k+1-a12k+3+a12k+5-a12k+7+a12k+9-a12k+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,T12n=-4n.                                       (3)解:T12m=-4m(m≥1).当n=12m+1,12m+2时,Tn=4m+1;当n=12m+3,12m+4时,Tn=-4m+1-r;当n=12m+5,12m+6时,Tn=4m+5-r;当n=12m+7,12m+8时,Tn=-4m-r;当n=12m+9,12m+10时,Tn=4m+4;当n=12m+11,12m+12时,Tn=-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.

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