来源：蜘蛛抓取(WebSpider)
时间：2011-09-29 09:48
标签：
b1 b2 b6
考点：数列与不等式的综合
专题：综合题,压轴题,等差数列与等比数列,不等式的解法及应用
分析：（1）直接由已知b3=3结合题目给出的数列递推式求b1的值；（2）由bn+2=-bn+1-bn，bn+3=-bn+2-bn+1，作差后得到bn+3=bn，进一步利用作差法证明数列{bnbn-1bn-2+n}是等差数列；（3）由Tn+1=Tn•bn+1=Tn-1bnbn+1=Tn-2bn-1bnbn+1=…=b1b2b3…bn+1，得到当n≥2时，Tn=b1b2b2…bn，然后推出数列{T3n-2+T3n-1+T3n）（n∈N*）是等比数列，求出其前n项和Sn=T1+T2+T3+…+Tn，再对n分当n=3k（k∈N*）时，当n=3k-1（k∈N*）时，当n=3k-2（k∈N*）时具体求和，最后利用放缩法得答案．
（1）解：∵bn+2=-bn+1-bn，∴b3=-b2-b1=-3b1=3，∴b1=-1；（2）证明：∵bn+2=-bn+1-bn ①，∴bn+3=-bn+2-bn+1 ②，②-①得bn+3=bn，∴（bn+1bn+2bn+3+n+1）-（bnbn+1bn+2+n）=bn+1bn+2（bn+3-bn）+1=1为常数，∴数列{bnbn+1bn+2+n}是等差数列；（3）解：∵Tn+1=Tn•bn+1=Tn-1bnbn+1=Tn-2bn-1bnbn+1=…=b1b2b3…bn+1，当n≥2时，Tn=b1b2b2…bn（*），当n=1时，T1=b1适合（*）式∴Tn=b1b2b3…bn（n∈N*）．∵b1=-12，b2=2b1=-1，b3=-3b1=32，bn+3=bn，∴T1=b1=-12，T2=T1b2=12，T3=T2b3=34，T4=T3b4=T3b1=34T1，T5=T4b5=T2b3b4b5=T2b1b2b3=34T2，T6=T5b6=T3b4b5b6=T3b1b2b3=34T3，…T3n+1+T3n+2+T3n+3=T3n-2b3n-1b3nb3n+1+T3n-1b3nb3n+1b3n+2+T3nb3n+1b3n+2b3n+3=T3n-2b1b2b3+T3n-1b1b2b3+T3nb1b2b3=34（T3n-2+T3n-1+T3n），∴数列{T3n-2+T3n-1+T3n）（n∈N*）是等比数列，首项T1+T2+T3=34且公比q=34，记Sn=T1+T2+T3+…+Tn，①当n=3k（k∈N*）时，Sn=（T1+T2+T3）+（T4+T5+T6）…+（T3k-2+T3k-1+T3k）=34[1-(34)k]1-34=3[1-(34)k]．∴34≤Sn＜3；②当n=3k-1（k∈N*）时，Sn=（T1+T2+T3）+（T4+T5+T6）+…+（T3k-2+T3k-1+T3k）-T3k=3[1-(34)k]-(b1b2b3)k=3-4•(34)k．∴0≤Sn＜3；③当n=3k-2（k∈N*）时，Sn=（T1+T2+T3）+（T4+T5+T6）+…+（T3k-2+T3k-1+T3k）-T3k-1-T3k=3[1-(34)k]-(b1b2b3)k-1b1b2-(b1b2b3)k=3[1-(34)k]-12(34)k-1-(34)k=3-143•(34)k．∴-12≤Sn＜3．综上得：-12≤Sn＜3．则p≤-12且q≥3．∴q-p的最小值为72．
点评：本题考查了等差关系的判断与应用，考查了等比关系的判断，训练了数列的分组求和与等比数列的前n项和，考查了数列不等式的解法，体现了分类讨论的数学思想方法，是压轴题．
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精英家教网新版app上线啦！用app只需扫描书本条形码就能找到作业，家长给孩子检查作业更省心，同学们作业对答案更方便，扫描上方二维码立刻安装！数列1/3,-1/8,1/15,-1/24……的通项公式an=?
蘇荷╱●qzul
观察可得3=1²+2×1 8=2²+2×2 15=3²+2×3 24＝4²+2×4所以an=1/（-1）n+1次（n²+2n）
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其他类似问题
通项公式an为-1的n+1次方乘以n（n+2）分子一。
(-1)^(n+1)/[n(n+2)] 设an的分母为bn，则b1=3, b2=3+5=8, b3=3+5+7=15, b4=3+5+7+9=24bn=3+5+7+9+...+(2n+1)=(3+2n+1)*n/2=(n+2)*nan的分子为(-1)^(n+1)an=(-1)^(n+1)/[n(n+2)]
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