mathematical inductiion, 一條功課…找不到一個簡單的方法計，請指點…

n^3 + 5n is divisible by 6 for all natural numbers n.


唔知有無邊位可以話我知有咩簡單的方法計到
我有個方法要寫長d，而且比較特別…所以徵求大家的指點

Let P(n) be the statement n^3 + 5n is divisible by 6 for all natural numbers n.

For n = 1
1^3 + 5(1) = 6(1)
n = 1 is true

Assume P(k) is true, i.e. k^3 + 5k = 6I, I = integer
To show P(k+1) is true,

(k+1)^3 + 5(k+1)
= k^3 + 3k^2 + 3k + 1 + 5k +5
= 6I + 3(k^2 + k + 2) (By assumption)

For k is odd number, k = 2m + 1, m = integer
6I + 3[(2m + 1)^2 + (2m + 1) + 2)]
= 6I + 3(4m^2 + 6m +4)
= 6( I + 2m^2 + 3m + 2)

For k is even number, k = 2m, m = integer
6I + 3(4m^2 + 2m + 2)
= 6( I + 2m^2 + m + 1)

Therefore, P(k+1) is true
By M.I., P(n) is true for all n.

本篇文章已被 鮑鮑 於 Sep 17 2005, 01:00 編輯過

Let P(n) be the proposition "n^3 + 5n is divisible by 6 for all natural numbers n."

when n = 1
1^3 + 5(1) = 6 = 6x1
∴P(1) is true

when n=2
2^3 + 5(2)=18
∴P(2) is true

Assume P(k) is true for all natural numbers ,
k^3 + 5k = 6M , M is a natural number
when n = k+2
(k+2)^3 + 5(k+2)
= k^3 + 4k^2 + 4k + 2k^2 + 8k + 8 + 5k + 10
= 6M + 6k^2 + 12k + 18
= 6(M + k^2 + 2k + 3)

∵(M + k^2 + 2k + 3) is a natural number
∴P(k+2) is true

by the principle of MI , P(n) is true for all natural numbers


這樣做可以嗎？？
我想盡量用一頁的位置做完這題…

I think it is already not very long. Why do you want to 用一頁的位置做完這題???

咁不如我幫你縮短喇=3=.......
照你做法縮短

唔駛寫第一句=3=

直接
for n=1 when n=2
1^3+5(1)=6 2^3 + 5(2)=18
=>S(1) is true =>S(2) is true

assume S(k), S(k-1) is true
i.e. k^3+5k=6M, for all M belongs to {N}
for n=k+1,
(學你三行做完, 唔做了=3=)
=6U , U belongs to {N}
it is true for n=k+1

BY MI, it is true for all n belongs to {N}

只是重新排版XDDD~
n=2寫上去n+1旁
另外跳step(老師唔會逐步睇的, 而且都冇step分=3=)
字再寫細D, 咁應該一張A5都寫得完= =+

其實你果個都夠少阿=3=...
一張紙絕對得

其實MI d題都係一般的長= ="
再少都大約要一頁紙的= ="

其實如果唔係為計而計, 呢題可以0甘理解:

n^3+5n的factors包括n(n^2+5)

有n呢個factor, 已經可以俾所有整數除得盡.
再加上6符合(n^2+5), when n=1, 就 "更加"除得盡.(remainder theorem)

其實要一個polynomial有整數factor有幾難, 多多都揾到.

本篇文章已被 pole 於 Dec 19 2005, 03:04 編輯過

這應該是 "for all n", n^3+5n can be divisible by ???,

也是不太明白你想說甚麼......-___-"

你自己試下求其代啲整數入去睇下, 睇下係咪淨係6先得? 梗唔係喇, 應該乜整數都得.
n(n^2+5) is divisible by n, n 又係啲乜?

remaimder theorem 適用於 real number, 整數係real number.
x(x^2+5)

n(n^2+5) 當然可以 divided by n,
那樣跟

1."你自己試下求其代啲整數入去睇下, 睇下係咪淨係6先得? 梗唔係喇, 應該乜整數都得."

2."n 又係啲乜?"

3."remaimder theorem 適用於 real number, 整數係real number."

4.這條題目

有甚麼關係??

同埋......

我真的不知你想說甚麼,你是想證明/說明些甚麼??可不可以說清楚一點?

本篇文章已被 阿暪 於 Dec 19 2005, 09:19 編輯過

你會不會有點理解錯誤?

是n=any real integer, f(n)都給6除得盡
不是給所有整數除得盡

或者這樣說
你只是用抽factor來証出:
when n=x, it is divisible by x (x is real number)
而不是
when n=x, it is divisible by 6

與題目是有明顯不同的

說一個類同而extreme的example
f(n) = n^2
是不是代表f(n) divisible by 6或者f(n) "俾所有整數除得盡"?
前者明顯不是
後者亦只有when n=x, it is divisible by x的必然關係
沒有一個數是"俾所有整數除得盡"

請指教

本篇文章已被 鮑鮑 於 Dec 19 2005, 09:25 編輯過

系 f(n)...... ,同n is integer...
↑啊,改了.

兩位繼續....

本篇文章已被 阿暪 於 Dec 19 2005, 09:21 編輯過

呀係喎, 唔單止要除得盡6, 仲要係代所有n都除得盡.
咁即係同(n^2+5)呢個factor有關啦.
sorry, sorry.
6, 9, 14, 21... 先得.

唔怪得阿瞞唔佑我嗡乜啦.

本篇文章已被 pole 於 Dec 19 2005, 12:41 編輯過