頁:
[1]
數學考試試題：Prove f(x)=n^3+5n is divisible by 6
考個時真系諗唔到，識做噶師兄麻煩幫下我，我想知道點樣做。<div></div> 本帖最後由 chongwaikei 於 2013-1-8 05:13 PM 編輯f(x) = n^3 + 5n < < < fail
you can prove by MI
or
n^3 + 5n = n^3 - n + 6n
= n(n - 1)(n + 1) + 6n
(n - 1) , n , (n + 1) are three different continuous numbers
so , n(n - 1)(n + 1) is divisible by 6
hence , f(n) = n^3 + 5n = n(n - 1)(n + 1) + 6n is divisible by 6
另外, 此乃台灣論壇, 請用書面語對話...
Could you please show me how to prove by MI.I fail to prove it by MI at that time. 190303458 發表於 2013-1-9 11:04 AM static/image/common/back.gif
Could you please show me how to prove by MI.I fail to prove it by MI at that time.
http://latex.codecogs.com/gif.latex?\\f(0)$%20is%20true%20$\\\\%20$assume%20when%20$n=k$%20the%20statement%20is%20true%20$\\\\%20$therefore%20,%20when%20$n=k+1\\\begin{align*}%20(k+1)^3+5(k+1)&=k^3+3k^2+3k+1+5k+5\\&=6m+3(k^2+k+2)\end{align*}\\%20$when%20$k$%20is%20odd%20,%20$k^2+k+2$%20is%20even$\\%20$when%20$k$%20is%20enev%20,%20$k^2+k+2$%20is%20even$\\\\%206m+3(k^2+k+2)=6m+6p=6(m+p)$%20where%20$m,p$%20are%20some%20+ve%20integers$\\\\%20$since%20,%20$f(k+1)$%20is%20true%20for%20$n\in\mathbb{Z}\setminus%20\mathbb{Z^-}%20\\\\%20$and%20$f(-n)=-f(n)$%20therefore%20,%20$f(n)$%20is%20true%20for%20negative%20integer$\\\\%20$by%20the%201st%20principle%20of%20MI%20,%20the%20statement%20is%20true%20for%20$n\in\mathbb{Z}
...<div class='locked'><em>瀏覽完整內容，請先 <a href='member.php?mod=register'>註冊</a> 或 <a href='javascript:;' onclick="lsSubmit()">登入會員</a></em></div>
頁:
[1]