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January 30th, 2019, 01:17 AM  #1 
Member Joined: Nov 2012 Posts: 80 Thanks: 1  Binomial theorem
How can I prove the equation by using binomial theorem? Thanks.

January 30th, 2019, 07:16 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
One way to start is to think $\displaystyle 2^{(n1)} = (1 + 1)^{(n1)} = \left (\sum_{j=0}^{n1} \dbinom{n1}{j} * 1^{(n1j)} * 1^j \right ) = \sum_{j=0}^{n1} \dbinom{n1}{j}.$ Now if n is even, I play around. If n = 2 $\dbinom{21}{0} + \dbinom{21}{1} = \dbinom{1}{0} + \dbinom{1}{1} = \dbinom{2 \div 2}{0} + \dbinom{2 \div 2}{1}.$ What if n = 4? $\dbinom{3}{0} + \dbinom{3}{1} + \dbinom{3}{2} + \dbinom{3}{3} = 1 + 3 + 3 + 1 = 8 =$ $1 + 6 + 1 = \dbinom{4}{0} + \dfrac{4 * 3}{2} + \dbinom{4}{4} = \dbinom{4}{0} + \dbinom{4}{2} + \dbinom{4}{4}.$ Hmm. Maybe $m \in \mathbb Z^+ \text { and } n = 2m \implies \displaystyle \sum_{j=0}^{n1} \dbinom{n1}{j} = \sum_{i=0}^{m}\dbinom{n}{2i}.$ Can you prove that? 
January 30th, 2019, 06:11 PM  #3 
Member Joined: Nov 2012 Posts: 80 Thanks: 1 
Thank you for your hints. My prove is as follow

January 31st, 2019, 12:39 AM  #4 
Senior Member Joined: Aug 2012 Posts: 2,332 Thanks: 723 
$\dbinom{n}{k} =$ the number of subsets of size k in a set of size n. Since the number of even and odd subsets of a finite set are equal, we're done. There are $2^n$ subsets, and the expression we're supposed to calculate just counts the number of even subsets, which is therefore $2^{n1}$. https://math.stackexchange.com/quest...areoddsized Last edited by Maschke; January 31st, 2019 at 12:51 AM. 

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