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#1 2006-07-03 01:02:19

mitochondria
Member
Registered: 2006-07-03
Posts: 3

Deriving an Expression for ∑nCr...

I've been trying to derive an expression for ∑nCr in the past hour and I got nothing out of it *frown*... I just kept expanding and cancelling the expansion using n!/[r!(n-r)!], I think I'm missing the pattern...

Thanks in advance!


P.S.  I couldn't find my high school text book tongue

Last edited by mitochondria (2006-07-03 12:37:11)

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#2 2006-07-03 01:36:11

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Deriving an Expression for ∑nCr...

With the summation being on r and n constant, right?  I believe it is 2^n.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2006-07-03 01:41:53

mitochondria
Member
Registered: 2006-07-03
Posts: 3

Re: Deriving an Expression for ∑nCr...

Yes tongue Sorry.. I should have mentioned that r = 0

Mm, that sounds familiar tongue Thank you very much.. I'll just go through my algebra and see what went wrong...

Just out of curiosity... n and r can vary =/? (Sorry.. I"m not a mathematician...)

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#4 2006-07-03 02:06:15

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Deriving an Expression for ∑nCr...

I should have mentioned that r = 0... Just out of curiosity... n and r can vary =/? (Sorry.. I"m not a mathematician...)

I don't really understand your question, and from them, it appears that you don't have too much experience with summations.  So just to make sure I cover all the bases, I'll go through a quick intro to summations, to make sure we are on the same page.

Summation is represented by:

Where i=0 is the starting point, and n is the ending point.  You can also replace n with a number:

In which case this sum (for example) would be 0 + 1 + 2 + 3 + 4 + 5 = 15.  The shorthand version for summation is:

Which means the summation from 0 to infinity of n.  This is very often used in calculus, and so this shorthand was adopted to make things easier to write.  Obviously, if you have 2 or more variables, you need to specify which one it is that you are changing.  So for example:

Doesn't make any sense while:

Does.  Or, you could do a double summation:

But be careful here, as sometimes the order of summation matters. 

Hopefully I covered your question somewhere in there.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2006-07-03 12:41:00

mitochondria
Member
Registered: 2006-07-03
Posts: 3

Re: Deriving an Expression for ∑nCr...

Mm... Thank you smile I do know about summation though, sorry about my rather obscure question.  What I was trying to say was that I forgot to specify that the limits are from r=0 to r=n at the beginning, that is:

I went through my algebra again and I just couldn't see how it gives 2^n (which is true for n is odd) *frown*...


Edited: I do know that it's just the summation of a row in the pascal triangle and I can prove it using the bionomial expression (x+y)^n where x = y = 1... I just can't make sense of the algebra and see how it gives 2^n tongue

Last edited by mitochondria (2006-07-03 13:02:14)

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#6 2006-07-06 02:30:12

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: Deriving an Expression for ∑nCr...

#5 right
(x+y)[sup]n[/sup]= [sub]n[/sub]C[sub]0[/sub]x[sup]n[/sup] +[sub]n[/sub]C[sub]1[/sub]x[sup]n-1[/sup]y +...+[sub]n[/sub]C[sub]n[/sub]y[sup]n[/sup]

let x=1 and y=1

2[sup]n[/sup]= the sum you want


X'(y-Xβ)=0

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#7 2006-07-06 05:36:34

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Deriving an Expression for ∑nCr...

What George posted is binomial theorm.  It took me a while to understand it even though I knew what he was trying to say!  So let me try to post cleaner version:

Binomial Theorm states that:

So if we let x = y = 1:


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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