pure maths summation

2009-10-10 4:57 am
想問 Σ(r+1)C 點變 Σ nC ??

p.s
黃色面:
Σ 上面係 (n-1) ,, 下面係 r=1
C 右上角係 n ,, 右下角係 (r+1)

藍色面:
Σ 上面係 (n-1) ,, 下面係 r=1
C 右上角係 (n-1) ,, 右下角係 r

plz explain steps by steps or teach me the technique used
really thz!!


回答 (2)

2009-10-10 5:36 am
✔ 最佳答案
要證: ∑[r=1,n-1](r+1)C(n,r+1) = ∑[r=1,n-1]nC(n-1,r)
(1 + x)^n = ∑[k=0,n]C(n,k)x^k
兩邊微分:
n(1 + x)^(n-1) = ∑[k=0,n]kC(n,k)x^(k-1)
n(1 + x)^(n-1) = ∑[k=1,n]kC(n,k)x^(k-1)
設k = r+1
n(1 + x)^(n-1) = ∑[r=0,n](r+1)C(n,r+1)x^r
n(1 + x)^(n-1) = C(n,1) + ∑[r=1,n](r+1)C(n,r+1)x^r
n(1 + x)^(n-1) = n + ∑[r=1,n](r+1)C(n,r+1)x^r
LHS = n(1 + x)^(n-1) = n∑[r=0,n-1]C(n-1,r)x^r
= nC(n-1,0) + n∑[r=1,n-1]C(n-1,r)x^r
= n + n∑[r=1,n-1]C(n-1,r)x^r
= RHS
所以n + ∑[r= 1,n](r+1)C(n,r+1)x^r = n + n∑[r=1,n-1]C(n-1,r)x^r
∑[r= 1,n](r+1)C(n,r+1)x^r = n∑[r=1,n-1]C(n-1,r)x^r
設x = 1
∑[r= 1,n](r+1)C(n,r+1) = n∑[r=1,n-1]C(n-1,r)得證

2009-10-09 21:40:20 補充:
更正:
上面所有∑[r= 1,n]都應該是∑[r= 1,n-1]

2009-10-09 21:43:03 補充:
http://img386.imageshack.us/img386/5332/sol.png
2009-10-10 5:28 am
may be this formula can help you :

(n+1)C(r+1)=nC(r+1)+nCr

2009-10-09 21:54:39 補充:
原來係我理解錯, 唔掛得做唔到


收錄日期: 2021-04-23 23:18:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091009000051KK01290

檢視 Wayback Machine 備份