an=2n-5,bn=an/2^n,设bn的前n项和为tn,证明;1/4大于等于tn小于1

发布网友发布时间：2024-10-09 07:05

共2个回答

热心网友时间：2024-10-09 21:37

题目有问题
bn=an/2^n
bn=(2n-5)/2^n
Tn=b1+b2+……+bn
=(2-5)/2+(2*2-5)/2^2+…+（2n-5）/2^n……............1
2Tn=(2-5)+(2*2-5)/2+(2*3-5)/2^2+……(2n-5)/2^(n-1) ................2
2式-1式得
Tn=-3+2/2+2/2^2+……+2/2^(n-1)-(2n-5)/2^n
=-3-(2n-5)/2^n+1+1/2+1/2^2+……+1/2^(n-2))
=-3-(2n-5)/2^n+[1-(1/2)^(n-1)]/(1-1/2)
=-3-(2n-5)/2^n+2[1-(1/2)^(n-1)]
=-3-(2n-5)/2^n+2-2*(1/2)^(n-1)
=-3-(2n-5)/2^n+2-4/2^n
=-1-(2n-5)/2^n-4/2^n
=-1-(2n-5+4)/2^n
=-1-(2n-1)/2^n

热心网友时间：2024-10-09 21:37

题可能有误。tn=-1-[(2n-1)/2^n]. 【1】易知，对任意n≥1,(2n-1)/2^n＞0.∴tn=-1-[(2n-1)/2^n]＜-1.【2】∵t(n+1)-tn=(2n-3)/2^(n+1).n=1,2,3,...t2-t1＜0.t2＜t3＜t4＜...＜tn.∴数列{tn}中，t2最小=-7/4.∴对任意n=1,2,3,..恒有-7/4≤tn＜-1.