解答

✨ 最佳解答 ✨

25.
a2 = a1 + f(0) = 2
f(0) = 2 - 1 = 1…為函數的常數項
a3 = a2 + f(1) = 5
f(1) = 5 - 2 = 3…為函數係數和
a4 = a3 + f(2) = 12
f(2) = 12 - 5 = 7…函數會通過(2, 7)
可假設函數為f(x) = y = a*x^2 + (3-1-a)x + 1
7 = 4a - 2a + 4 + 1
=> 2a = 2
=> a = 1
f(x) = y = x^2 + x + 1
a5 = a4 + f(3) = 12 + 9 + 3 + 1 = 25

38.
令三數為a ar a*r^2
a + ar + a*r^2 = a(r^2 + r + 1) = 39…①
a^2 + (ar)^2 + (a*r^2)^2
= (a^2)(r^4 + r^2 + 1)
= (a^2)(r^2 + r + 1)(r^2 - r + 1)
= [a(r^2 + r + 1)][a(r^2 - r + 1)]
= 819

a(r^2 - r + 1) = 819/39 = 21…②
兩式相減(①-②)得
2ar = 18
ar = 9

代回原式①得
9/r + 9 + 9r = 39
=> 3*r^2 - 10r + 3 = 0
=> (3r-1)(r-3) = 0
=> r = 1/3 or 3
無論r是哪個解
都會得出三數由小到大為3 9 27

原來是要這樣算!謝謝你~~

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