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Pn = k*P(n-1) + h*P(n-2) P2 = k*P1 + h*Po = 9k + 8h = 57 P3 = k*P2 + h*P1 = 57k + 9h = 111 => k = 1, h = 6 => Pn = P(n-1) + 6*P(n-2) Characteristic equation: 0= x^2 - x - 6 = (x-3)(x+2) => x = 3, -2 => Pn = p*3^n + q*(-2)^n Po = p + q = 8 P1 = 3p - 2q = 9 => p = 5, q = 3 => Pn = 5*3^n + 3*(-2)^n => P2 = 5*9 + 3*4 = 57 => P3 = 5*27 - 3*8 = 111 => P4 = 5*81 + 3*16 = 453 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 118.169.46.6 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1426021488.A.D8F.html
yyc2008 : 請問怎麼確定k h是定值? 03/11 15:16
ctchang34 : Po.P1.P2.P3代入解聯立方程式 03/12 14:41