请看图,有关求化简的, 求数学大神解答,好的解答有好评,要有过程,谢谢。求x的集合?
1个回答
2016-02-15
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(Ⅰ)由题意得:a1=2000(1+50%)-d=3000-d,
a2=a1(1+50%)-d=3/2a1-d=4500-5/2d,…
an+1=an(1+50%)-d=3/2an-d.
(Ⅱ)由(Ⅰ)得an=3/2a(n-1)-d=3/2(3/2a(n-2)-d)-d=(3/2)^2 a[n-2]-3/2d-d
=…
=(3/2)^(n-1)a1-d[1+3/2+(3/2)^2+…+(3/2)^(n-2)]
整理得:an=(3/2)^(n-1)(3000-d)-2d[(3/2)^(n-1)-1]
=(3/2)^(n-1)(3000-3d)+2d.
由题意,am=4000,即(3/2)^(m-1)(3000-3d)+2d=4000.
解得d={[(3/2)^m-2] ×1000}/[(3/2)^m-1]
=1000(3^m-2^(m+1))/[3^m-2^m],
故该企业每年上缴资金d的值为1000(3^m-2^(m+1))/[3^m-2^m]时,经过m(m≥3)年企业的剩余资金为4000万元.
a2=a1(1+50%)-d=3/2a1-d=4500-5/2d,…
an+1=an(1+50%)-d=3/2an-d.
(Ⅱ)由(Ⅰ)得an=3/2a(n-1)-d=3/2(3/2a(n-2)-d)-d=(3/2)^2 a[n-2]-3/2d-d
=…
=(3/2)^(n-1)a1-d[1+3/2+(3/2)^2+…+(3/2)^(n-2)]
整理得:an=(3/2)^(n-1)(3000-d)-2d[(3/2)^(n-1)-1]
=(3/2)^(n-1)(3000-3d)+2d.
由题意,am=4000,即(3/2)^(m-1)(3000-3d)+2d=4000.
解得d={[(3/2)^m-2] ×1000}/[(3/2)^m-1]
=1000(3^m-2^(m+1))/[3^m-2^m],
故该企业每年上缴资金d的值为1000(3^m-2^(m+1))/[3^m-2^m]时,经过m(m≥3)年企业的剩余资金为4000万元.
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