Re: [中學] 餘式定理
: b題不太會
: 請幫一下
假設沒學過二項式定理
n = 2k + 1
f(x) = x^(2k + 1) + 2x + 3
= (x - 1)[1 + x + x^2 + ... + x^(2k)] + 1 + 2x + 3
= (x - 1) + (x - 1)(x + 1)[x + x^3 + ... + x^(2k - 1)] + 2x + 4
= (x^2 - 1)[x + x^3 + ... + x^(2k - 1)] + (3x + 3)
Remainder
如果必須用到(a)的結果
f(x) = (x + 1)Q(x)
= x^(2k + 1) + 2(x + 1) + 1
=> (x + 1)[Q(x) - 2] = x^(2k + 1) + 1
= (x + 1)[1 - x + x^2 - ... + x^(2k)]
=> f(x) = (x + 1)[3 - x + x^2 - ... + x^(2k)]
= 3(x + 1) + (x + 1)(x - 1)[x + x^3 + ... + x^(2k - 1)]
Remainder
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※ 編輯: Honor1984 (111.249.168.201), 06/10/2017 22:19:21
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