pure maths

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2009-11-22 5:44 am

圖片參考：http://imgcld.yimg.com/8/n/HA00221188/o/700911210146913873409640.jpg

p.s. it is noted that there are some information of the qurestion at the top right hand corner of the above picture .

here are the details: where S(n+1) = a1 + a2 +...+a(n+1)

thz a lot

回答 (1)

用戶9112

2009-11-22 7:38 am

✔ 最佳答案

(a) 1/S(n+1) - 1 / S(n)
= [S(n) - S(n+1)]/[S(n+1)S(n)]
= [S(n+1) - a(n+1) - S(n+1)] / {S(n+1)[S(n+1) - a(n+1)]}
= -a(n+1) / [S^2(n+1) - S(n+1)a(n+1)]
Since a(n+1) = 2S^2(n+1) / [2S(n+1) - 1]
S^2(n+1) = a(n+1)[2S(n+1) - 1]/2
1/S(n+1) - 1 / S(n)
= -a(n+1) / {a(n+1)[2S(n+1) - 1]/2 - S(n+1)a(n+1)}
= -a(n+1) / [-a(n+1)/2]
= 2 is a constant
Therefore 1/S(1), 1/S(2), 1/S(3), ... is an arithmetic sequence
(b) Since S(1) = 2, the first term of the AM is 1/2, the common diference is 2
1/S(n) = 1/2 + (n-1)(2) = 2n - 3/2 = (4n - 3)/2
S(n) = 2/(4n-3)
S(n-1) = 2/[4(n-1) - 3] = 2/(4n-7)
a(n) = S(n) - S(n-1)
= 2/(4n-3) - 2/(4n-7)
= -8/[(4n-3)(4n-7)]
lim n-> ∞ n^2a(n)
= lim n-> ∞ -8n^2/[(4n-3)(4n-7)]
= lim n-> ∞ -8/[(4-3/n)(4-7/n)]
= -8/16
= -1/2

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