推 cnick :很感謝你的幫忙 我了解了 12/26 20:59
※ 引述《cnick (cnick)》之銘言：
: 各位好
: 有一題數列求和問題想請教大家
: ___________________________________________________________________________
: 已知 a_1=2,點 (a_n,a_(n+1) ) 在函數f(x)=x^2+2x的圖像上,其中 n=1,2,3,…
: 設 b_n=1/a_n +1/(a_n + 2 ), 試求數列{b_n}的前n項和S_n
: 解答: 1-2/(3^(2^n )-1)
: ____________________________________________________________________________
: 因為符號在這有點難打
: 所以也可以請看下列網址的word檔
: https://www.asuswebstorage.com/navigate/s/3247710444F74B42AB996BE60FD8196FY
: 謝謝大家的回答
亦即a_(n+1) = (a_n)^2+2(a_n) = (a_n)(a_n +2)
倒數得1/a_(n+1) = 1/(a_n)(a_n +2) =[1/a_n - 1/(a_n +2)](1/2)
兩邊減1/a_n得1/a_(n+1) - 1/a_n = (-1/2)[1/a_n + 1/(a_n +2)] = (-1/2)(b_n)
則b_n = -2[1/a_(n+1) - 1/a_n]
b_(n-1)=-2[1/a_n - 1/a_(n-1)]
...
b_1 = -2[1/a_2 - 1/a_1]
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加總得S_n = (-2)[1/a_(n+1) - 1/a_1]
又a_(n+1) = (a_n)^2+2(a_n) = (a_n +1)^2-1
即a_(n+1)+1 =(a_n +1)^2 = [(a_(n-1)+1)^2]^2 = [a_(n-1)+1]^4
= ... = (a_1 +1)^(2^n) = 3^(2^n)
故S_n = (-2)[1/(3^(2^n)-1) - 1/2] = 1 -2/(3^(2n)-1)
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