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

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/04 07:09:35

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

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