f.4 m.i. please come and help

網路上的路人甲

2010-12-14 3:33 am

Prove, by mathematical induction, that x^(n+2) + (x+1)^(2n+1) is divisible by (x^2) + x + 1 for all positive integers n.

更新1:

請問可唔可以解下 x * x^(k+2) + (x^2 + 2x + 1) * (x+1)^(2k+1) 點變做 x * x^(k+2) + x * (x+1)^(2k+1) + (x^2 + x + 1) * (x+1)^(2k+1) 我唔明點解將(x+1)^2變做x^2+2x+1後之後果行變左做x * x^(k+2) + x * (x+1)^(2k+1) + (x^2 + x + 1) * (x+1)^(2k+1) thz

回答 (1)

用戶2053

2010-12-14 3:55 am

✔ 最佳答案

When n = 1 ,x^3 + (x+1)^3 = (2x + 1)[x^2 - x(x+1) + (x+1)^2]= (2x + 1)(x^2 + x + 1) is true ,Assume that when n = k the statement is true :x^(k+2) + (x+1)^(2k+1) = m(x^2 + x + 1)When n = k+1 :x^(k+1 +2) + (x+1)^(2(k+1) + 1)= x^(k+3) + (x+1)^(2k+3) = x * x^(k+2) + [(x+1)^2] * (x+1)^(2k+1) = x * x^(k+2) + (x^2 + 2x + 1) * (x+1)^(2k+1) = x * x^(k+2) + x * (x+1)^(2k+1) + (x^2 + x + 1) * (x+1)^(2k+1) = x * [x^(k+2) + (x+1)^(2k+1)] + (x^2 + x + 1) * (x+1)^(2k+1)= x * m(x^2 + x + 1) + (x^2 + x + 1) * (x+1)^(2k+1)= (x^2 + x + 1) [mx + (x+1)^(2k+1)] is divisible by x^2 + x + 1.By mathematical induction it is true for all positive integers.

2010-12-13 20:15:32 補充：
x * x^(k+2) + (x^2 + 2x + 1) * (x+1)^(2k+1)

好簡單 :

將 (x^2 + 2x + 1) 拆成 (x^2 + x + 1) + x ,

於是
= x * x^(k+2) + [(x^2 + x + 1) + x] * (x+1)^(2k+1)

= x * x^(k+2) + (x^2 + x + 1)(x+1)^(2k+1) + x * (x+1)^(2k+1)

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