作者yueayase (scrya)
看板Math
标题Re: [中学] 无穷等比级数
时间Wed Apr 19 02:49:38 2023
※ 引述《ChenYM (老宅男一个)》之铭言：
: https://i.imgur.com/0MJA3K7.jpg
: 这一题该怎麽着手？
: 请前辈赐教
令a 公比r b 公比r
n 1 n 2
两个首项都为1 => a = b = 1
1 1
=>Σa = 1/(1-r ), Σb = 1/(1-r )
n 1 n 2
因为a 和b 都为等比数列，所以a b 也是等比数列，且首项为1*1=，1公比为r r
n n n n 1 2
=>Σa b = 1/(1-r r )
n n 1 2
由题目设定，
1/(1-r r ) = 3/4 => 1-r r = 4/3 => r r = -1/3
1 2 1 2 1 2
Σ(a + b ) = Σa + Σb = 1/(1-r ) + 1/(1-r ) = 13/5
n n n n 1 2
(1-r ) +(1-r ) (2-r -r ) 2 - r - r
2 1 1 2 1 2
=> -------------- = ------------ = ------------- = 13/5
(1-r )(1-r ) 1-r -r +r r 2/3 -r -r
1 2 1 2 1 2 1 2
=> 5(2-r - r ) = 13(2/3-r - r )
1 2 1 2
=> 15[2-(r +r )] = 13[2-3(r + r )]
1 2 1 2
=> 24(r + r ) = -4
1 2
=> r + r = -1/6
1 2
2 2
=> (r -r ) = (r +r ) -4r r = 1/36 + 4/3 = 49/36
1 2 1 2 1 2
=> r -r = ±7/6
1 2
若r - r = 7/6 => 2r = 1 => r = 1/2, r = -1/6 - 1/2 = -4/6 = -2/3
1 2 1 1 2
若r - r = -7/6 => 2r = -8/6 => r = -2/3, r = -1/6 + 2/3 = 3/6 = 1/2
1 2 1 1 2
=> 两公比为1/2和-2/3
2 2
因为a 和 b 为等比数列，所以a 和b 也为等比数列
n n n n
2 2 2
且首项皆为1 = 1，公比各自为r 和r
1 2
2 2 2 2 2
∴Σ(a + b ) = Σa +2Σa b +Σb = 1/(1-r ) + 2*3/4 + 1/(1-r )
n n n n n n 1 2
= 1/(1-1/4) + 3/2 + 1/(1-4/9)
= 4/3 + 3/2 + 9/5
= 40/30 + 45/30 + 54/30
= 139/30
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