西雅圖一戰 (v34) 貢獻新閱讀狗兒 2 篇 (應該是新的)

dianewei525

這題我也想了很久 我是把 (n+1)!拆開變成 (n+1)(n)(n-1)! 帶進去formula裡面
1. (n+1)!n!+(n-1)!n!= n(n+1)!(an^2+bn+c), what's abc?

soln:
(n+1)(n)n!(n-1)!+(n-1)!n!= ((n+1)(n)+1)(n-1)!(n)!=(n^2+n+1)(n-1)!(n)!  <-higlight 的地方是原本的(n+1)! ..順序寫的有點亂有點confusing 不好意思
a=1, b=1, c=1; abc=1

dianewei525

another math JJ, it's real easy though
1. which of the following set of 3 numbers fit the criteria below:
a) any two out of the three numbers in the set  have a common factor >1
b) the three numbers together does not share any common factor >1

answer: 24, 28, 63

dianewei525

zzhong, haha thx for your correction >///<

lg09nnw

楼主好人，请问碰到的数学jj多嘛

forgmat11

1.maybe timo is: (n+1)!n!+(n-1)!n!= n(n-1)!(an^2+bn+c), what's abc?,
not  (n+1)!n!+(n-1)!n!= n(n+1)!(an^2+bn+c), what's abc?