OK2013文科数学数列专题练习(3)

(2) Tn?a1a2?an???4???5?n2?n2 ,

?4?Tn?1?a1a2?an?1????5?(n?1)2?(n?1)2(n?2),

T?4??an?n???Tn?1?5?n?1(n?2),

n?1?4?又a1?T1?1满足上式. 所以an????5?(n?N?)……………7分

(3) 若5f(an)是bn与an的等差中项, 则2?5f(an)?bn?an,

从而10(an2?12139an)?bn?an, 得bn?5an2?6an?5(an?)2?. 255n?1?4?因为an????5?当an?(n?N?)是n的减函数, 所以

3, 即n?3(n?N?)时, bn随n的增大而减小, 此时最小值为b3; 53当an?, 即n?4(n?N?)时, bn随n的增大而增大, 此时最小值为b4.

5又a3?33?a4?, 所以b3?b4, 5522??4?2?224?4?即数列{bn}中b3最小, 且b3?5?????6????. …………12分

125?5???5????15. 解：（Ⅰ）由题可得f'(x)?2x．

所以曲线y?f(x)在点(xn,f(xn))处的切线方程是：y?f(xn)?f'(xn)(x?xn)．

2即y?(xn?4)?2xn(x?xn)．

2令y?0，得?(xn?4)?2xn(xn?1?xn)． 2即xn?4?2xnxn?1．

显然xn?0，∴xn?1?xn2?． 2xnxn2(xn?2)2(xn?2)2xn2?，知xn?1?2???2?（Ⅱ）由xn?1?，同理xn?1?2?． 2xn2xn2xn2xnxn?1?2x?22x?2x?2?(n)．从而lgn?1?2lgn 故，即an?1?2an．所以，数列{an}成等比数列．故

xn?1?2xn?2xn?1?2xn?2x?2x?2an?2n?1a1?2n?1lg1?2n?1lg3．即lgn?2n?1lg3．

x1?2xn?2从而

xn?2?32所以xn?2n?1 xn?23?1n?12(32?1)n?1 11

（Ⅲ）由（Ⅱ）知xn?2(332n?1n?1?1)?1n?14bn?132?11111?0∴bn?xn?2?n?1∴ ?????nn?1n?11?122222bn33?13?13?133111当n?1时，显然T1?b1?2?3．当n?1时，bn?bn?1?()2bn?2???()n?1b1

3331b1[1?()n]1113∴Tn?b1?b2???bn?b1?b1???()n?1b1??3?3?()n?3．

13331?3 综上，Tn?3(n?N*)．

2，

12

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