Re: [微積] 用柯西證收斂

[JohnMash (Paul)]

設數列 〈An〉 滿足

 A_(n+1) = 1 / (A_n) +1     (分母是 A_n  + 1)

n是正整數

A_1 = 6

求

A_2011 - A_2010 等於多少



          Let a_n + a_{n+1} = a_{n+2},  n=0,1,2,3,....
          and a0=6, a1=1

  then

 a0=6, a1=1, a2=7, a3=8, a4=15, a5=23, a6=38, a7=61,....

  then

          A_n = a_{n-1}/a_n

  we can easily prove that

          (a_k)^2 - a_{k-1} * a_{k+1} = 41 for k is even
                                       -41 for k is odd


   A_{n+1} - A_n =  (a_n^2 - a_{n-1}*a_{n+1})/(a_n * a_{n+1})

                 = 41/(a_n*a_{n+1}) for n is even

                 = -41/(a_n*a_{n+1}) for n is odd

   however, f1=1, f2=1, f3=2, f4=3, f5=5, f6=8, ....

   denote b_n = a_n - f_n

   then b1=0, b2=6, b3=6, b4=12, b5=18,.....

   that is, b_n = 6 f_{n-1}

   hence, a_n = 6 f_{n-1} + f_n



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※ 編輯: JohnMash        來自: 27.147.57.77         (08/28 03:00)
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