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December 3rd, 2018, 08:28 AM   #1
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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 12:31 PM.
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December 3rd, 2018, 08:37 AM   #2
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Quote:
Originally Posted by mathbeginner2018 View Post
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.
Please do not double post. Someone will get to the problem. Just be a little patient.

-Dan

Last edited by skipjack; December 3rd, 2018 at 11:51 AM.
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December 3rd, 2018, 09:37 AM   #3
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Have you tried a double inductive proof?
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December 3rd, 2018, 09:42 AM   #4
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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 !
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December 3rd, 2018, 11:00 AM   #5
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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 $

$\displaystyle \left ( \sum_{k=0}^n \dbinom{m + k}{k} * 2^{(n-k)} \right ) + \left ( \sum_{k=0}^m \dbinom{n + k}{k} * 2^{(m-k)} \right ) = 2^{(m+n+1)}.$

$\displaystyle n = 1 = m \implies \left ( \sum_{k=0}^n \dbinom{m + k}{k} * 2^{(n-k)} \right ) + \left ( \sum_{k=0}^m \dbinom{n + k}{k} * 2^{(m-k)} \right ) =$

$\displaystyle \left ( \sum_{k=0}^1 \dbinom{1 + k}{k} * 2^{(1-k)} \right ) + \left ( \sum_{k=0}^1 \dbinom{1 + k}{k} * 2^{(1-k)} \right ) =$

$\left ( \dbinom{1}{0} * 2^{(1-0)} \right) \left ( \dbinom{2}{1} * 2^{(1-1)} \right) + \left ( \dbinom{1}{0} * 2^{(1-0)} \right) \left ( \dbinom{2}{1} * 2^{(1-1)} \right) =$

$1 * 2 + 2 * 1 + 1 *2 + 2 * 1 = 2 + 2 + 2 + 2 = 8 = 2^{(1 + 1 + 1)}. \text {In short,}$

$\displaystyle n = 1 = m \implies \left ( \sum_{k=0}^n \dbinom{m + k}{k} * 2^{(n-k)} \right ) + \left ( \sum_{k=0}^m \dbinom{n + k}{k} * 2^{(m-k)} \right ) = 2^{(n + m + 1)}.$

Now what?
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December 3rd, 2018, 11:48 AM   #6
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JeffM1,Thank you for the fast answer.

The equation is correct when n=m=1, but this does not prove anything. Anyways thank you!
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December 3rd, 2018, 12:06 PM   #7
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Quote:
Originally Posted by mathbeginner2018 View Post
JeffM1,Thank you for the fast answer.

The equation is correct when n=m=1, but this does not prove anything. Anyways thank you!
Of course it proves something. It is the start of an inductive proof.
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December 3rd, 2018, 12:31 PM   #8
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Quote:
Originally Posted by JeffM1 View Post
Of course it proves something. It is the start of an inductive proof.
Okay, you are right, it should be a right approach, I tried prooving it. But I didn't manage to do it. Thank you very much anyways.
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December 3rd, 2018, 12:45 PM   #9
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Quote:
Originally Posted by JeffM1 View Post
Of course it proves something. It is the start of an inductive proof.
https://www.wolframalpha.com/input/?...rom+0+to+n%2B1


I got here, that is absolutely correct. I jut cannot figure out how I will get this result.
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