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1 Answers · Science & Mathematics · 24/05/2010

**k*****C**(n+1,k) = k * (n+1)!/(k!(n+1-k)!) = (n+1)!/((k-1)!(n+1-k)!) = (n+1) * n!/((k-1)!(n+1-k)!) = (n+1) * C(n, k-1)1 Answers · Science & Mathematics · 14/08/2007

Use the Quotient Rule. = [1-e^(-alpha(u)] [

**k**+(**c**)(lamba)] - [**k**+(**c**)(lamda)(u)) (-e^(-alpha(u)) (-alpha)] / (1-e^(-alpha)(u)]^2 Note...1 Answers · Science & Mathematics · 27/03/2008

n * C(n-1, k-1) = n * [(n-1)! / ((k-1)! (n-k)!)] = n! / ((k-1)! (n-k)!) = k * [n! / (k! (n-k)!)] =

**k*****C**(n, k). I hope this helps!1 Answers · Science & Mathematics · 20/11/2011

Note that c_

**k**-**c**_{k-1} = 2k. The sum of (c_**k**-**c**_{k-1}) from k=1 to n is (c_n - c...3 Answers · Science & Mathematics · 21/03/2008

4x + 7 - 9x = 9 4x - 9x = 9 - 7 -5x = 2 x = -2/5 = -0.4

1 Answers · Science & Mathematics · 16/04/2011

... A and B are in

**K**(**C**) Xt (A+B) X = (... C X^N will be in**K**(**C**) by induction. But you can realise...2 Answers · Science & Mathematics · 22/10/2007

1) if log

**k**=**c**log v + log p, what is the value of k? start with the RHS: two ...2 Answers · Science & Mathematics · 31/03/2008

ƒ(k) = √[k + (

**k**−**c**)^2], assuming c is a constant, and k is the only variable. ƒ...v, by chain rule, we get: ƒ'(k) = u' ⋅ v ⋅ u^(v - 1) ƒ'(k) = [1 + 2(**k**−**c**)] ⋅ 1/2 ⋅ [k + (**k**−**c**)^2]^(-1/2) ƒ'(k) = (1 + 2k − 2c) / (2√[k + (**k**−**c**...3 Answers · Science & Mathematics · 23/02/2013

Pascal's rule: C(n,

**k**) =**C**(n-1,**k**) +**C**(n-1, k-1) First, show...m!n!) out the window to get sum [k=0 to n] (-1)^**k*****C**(n, k) = 0 which again can be shown easily by...2 Answers · Science & Mathematics · 30/07/2009