Proving the Equality of Newton Binomial Coefficient Using the Summation Method

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The discussion revolves around proving the equality of the Newton binomial coefficient using the summation method, specifically showing that the sum from k=0 to n of (n over k) equals 2^n. Participants suggest using the binomial theorem, with the recommendation to set x and y to 1 to simplify the proof. There is some confusion about the variables in the binomial theorem, particularly regarding their values and roles. The conversation emphasizes the importance of understanding the theorem before proceeding with the proof. Ultimately, the focus remains on clarifying the application of the binomial theorem to achieve the desired equality.
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Homework Statement



http://img82.imageshack.us/img82/8125/78492134fy0.th.jpg http://g.imageshack.us/thpix.php

I need to prove that the left part is equal to the right. I'm not sure how to approach the question.
I know that (n over k)=n! : k!(n-k)! but how do I sum all the number from k=0 to k=n and show that that equals to 2^n ?
 
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Abukadu said:

Homework Statement



http://img82.imageshack.us/img82/8125/78492134fy0.th.jpg http://g.imageshack.us/thpix.php

I need to prove that the left part is equal to the right. I'm not sure how to approach the question.
I know that (n over k)=n! : k!(n-k)! but how do I sum all the number from k=0 to k=n and show that that equals to 2^n ?

use the binomial theorem and let x=y=1

marlon
 
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hi marlon
what do you mean by x=y=1?
what is my x and y? the binomial theorem has r and y
 
What IS the binomial theorem?
 
We were kind of hoping that Abukadu would look it up himself! And, no, y is NOT equal to 0. If it were you would just have xn= xn
 
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