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December 3rd, 2018, 07:28 AM  #1 
Newbie Joined: Dec 2018 From: england Posts: 7 Thanks: 0 
Hello, I am new to the forum, so hello to everyone. I need some help with 2 equations that need to be proved. Thank you in advance. Last edited by skipjack; December 3rd, 2018 at 11:31 AM. 
December 3rd, 2018, 07:37 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,266 Thanks: 933 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan Last edited by skipjack; December 3rd, 2018 at 10:51 AM.  
December 3rd, 2018, 08:37 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
Have you tried a double inductive proof?

December 3rd, 2018, 08:42 AM  #4 
Newbie Joined: Dec 2018 From: england Posts: 7 Thanks: 0 
I haven't, but I have some instructions: For the first equation you can use combinational reasoning. For the 2nd equation you can use binomial formula and the multinomial formula. Use both the binomial and the multinomial formula for the following expression : ( 1 + u^2 + 1 / (4(u^2)) ) ^ (2n) Both things we get are equal ! 
December 3rd, 2018, 10:00 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
Are you required to use those methods? Because at least for the first, I find the inductive approach very easy to see although tedious in the mechanics. Probably the suggested method is very quick to do once you "see" it. $\text {PROVE: } n,\ m \in \mathbb Z^+ \implies $ $\displaystyle \left ( \sum_{k=0}^n \dbinom{m + k}{k} * 2^{(nk)} \right ) + \left ( \sum_{k=0}^m \dbinom{n + k}{k} * 2^{(mk)} \right ) = 2^{(m+n+1)}.$ $\displaystyle n = 1 = m \implies \left ( \sum_{k=0}^n \dbinom{m + k}{k} * 2^{(nk)} \right ) + \left ( \sum_{k=0}^m \dbinom{n + k}{k} * 2^{(mk)} \right ) =$ $\displaystyle \left ( \sum_{k=0}^1 \dbinom{1 + k}{k} * 2^{(1k)} \right ) + \left ( \sum_{k=0}^1 \dbinom{1 + k}{k} * 2^{(1k)} \right ) =$ $\left ( \dbinom{1}{0} * 2^{(10)} \right) \left ( \dbinom{2}{1} * 2^{(11)} \right) + \left ( \dbinom{1}{0} * 2^{(10)} \right) \left ( \dbinom{2}{1} * 2^{(11)} \right) =$ $1 * 2 + 2 * 1 + 1 *2 + 2 * 1 = 2 + 2 + 2 + 2 = 8 = 2^{(1 + 1 + 1)}. \text {In short,}$ $\displaystyle n = 1 = m \implies \left ( \sum_{k=0}^n \dbinom{m + k}{k} * 2^{(nk)} \right ) + \left ( \sum_{k=0}^m \dbinom{n + k}{k} * 2^{(mk)} \right ) = 2^{(n + m + 1)}.$ Now what? 
December 3rd, 2018, 10:48 AM  #6 
Newbie Joined: Dec 2018 From: england Posts: 7 Thanks: 0 
JeffM1,Thank you for the fast answer. The equation is correct when n=m=1, but this does not prove anything. Anyways thank you! 
December 3rd, 2018, 11:06 AM  #7 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  
December 3rd, 2018, 11:31 AM  #8 
Newbie Joined: Dec 2018 From: england Posts: 7 Thanks: 0  
December 3rd, 2018, 11:45 AM  #9  
Newbie Joined: Dec 2018 From: england Posts: 7 Thanks: 0  Quote:
I got here, that is absolutely correct. I jut cannot figure out how I will get this result.  

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