※ 引述《oxs77 (安)》之銘言：
: A,B,C,D,E 為等差數列
: A^3+6C^3+E^3
: 求 ------------ =
: B^3+D^3
a, e 滿足 U^2 - 2cU + (c^2-4t^2) = 0 , t:公差
b, d 滿足 V^2 - 2cV + (c^2-t^2) =0.
Let S_i=a^i+e^i - 4b^i-4d^i
Then S_(n+2) - 2cS_(n+1) + c^2S_n = 0.
S_(n+2) - cS_(n+1) = c(S_(n+1)-cS_n).
Note that S_0=-6, S_1=-6c ==> S_n = -6c^n.
Thus, (a^n+6c^n+e^n)/(b^n+d^n) = 4, if c!=0, for all n>=1.
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◆ From: 76.94.119.209
※ 編輯: Sfly 來自: 76.94.119.209 (03/05 17:19)
※ 編輯: Sfly 來自: 76.94.119.209 (03/05 17:39)