Binomial (1 Viewer)

=)(= · Oct 19, 2023

does anyone know how to do this?

Sam14113 · Oct 19, 2023

=)(= said:
View attachment 40745
does anyone know how to do this?

Here’s how I would do it:
- Substitute x=1 into the given equation
- Then on the LHS, invoke the symmetry property of Pascal’s Triangle (m choose k = m choose m-k for all appropriate m and k) on only the terms of the form 2n choose k, where k>n.
- Then add 2n choose n to both sides and divide both sides by 2

Unless Drongoski is correct, this should get you to the solution

scaryshark09 · Oct 19, 2023

Drongoski said:
I think the formula is incorrect. As a counter-example: let n = 3, x = 1

$(1+1)^{2x3} = 2^6 = 64\\ \\ 2^{2xn-1} = 2^5 = 32$

Hold it; maybe I'm wrong!

It only goes to n on the second one, whereas the first one goes to 2n, hence why it’s double the answer

Sam14113 · Oct 19, 2023

Drongoski said:
I think the formula is incorrect. As a counter-example: let n = 3, x = 1

$(1+1)^{2x3} = 2^6 = 64\\ \\ 2^{2xn-1} = 2^5 = 32$

Hold it; maybe I'm wrong!

Can you elaborate? Not sure which equation you’re saying is wrong and which sides you’re substituting into

Sam14113 · Oct 19, 2023

scaryshark09 said:
It only goes to n on the second one, whereas the first one goes to 2n, hence why it’s double the answer

Oh nvm got it

Drongoski · Oct 19, 2023

Proof:

$2^ {2n} = (1+1)^{2n} = \left [ \binom {2n} 0 + \binom {2n} 1 + \binom {2n} 2 + \cdots \binom {2n} n \right ] \\ \\ + \left [\binom {2n} n + \binom {2n} {n+1} + \binom {2n} {n+2} + \cdots + \binom {2n}{2n} \right] - \binom {2n} n \\ \\ = 2 \times \left [ \binom {2n} 0 + \binom {2n} 1 + \binom {2n} 2 + \cdots \binom {2n} n \right ] - \frac {(2n)!}{n!\cdot n!}\\ \\ \therefore 2 \times \left[\binom {2n} 0 + \binom {2n} 1 + \binom {2n} 2 + \cdots + \binom {2n} n \right ] = 2^{2n} + \frac {(2n)!}{(n!)^2} \\ \\ \therefore \binom {2n} 0 + \binom {2n} 1 + \binom {2n} 2 + \cdots + \binom {2n} n = 2^{2n-1} + \frac {(2n)!}{2\times (n!)^2}$

Retrodawn · Oct 19, 2023

Drongoski said:
Proof:

$2^ {2n} = (1+1)^{2n} = \left [ \binom {2n} 0 + \binom {2n} 1 + \binom {2n} 2 + \cdots \binom {2n} n \right ] \\ \\ + \left [\binom {2n} n + \binom {2n} {n+1} + \binom {2n} {n+2} + \cdots + \binom {2n}{2n} \right] - \binom {2n} n \\ \\ = 2 \times \left [ \binom {2n} 0 + \binom {2n} 1 + \binom {2n} 2 + \cdots \binom {2n} n \right ] - \frac {(2n)!}{n!\cdot n!}$

Hence the result (too tedious to type out the rest!)

wait how do u minus the 2ncn, and still factor out a 2?

scaryshark09 · Oct 19, 2023

Retrodawn said:
wait how do u minus the 2ncn, and still factor out a 2?

Cause he counted it twice

Retrodawn · Oct 19, 2023

scaryshark09 said:
Cause he counted it twice

ye so he has one left in his expansion, so how does he end up with 2 again?

scaryshark09 · Oct 19, 2023

Retrodawn said:
ye so he has one left in his expansion, so how does he end up with 2 again?

He cancels out the double up, and then he has 1 of each left from 1 to 2n
Since this is symmetrical, he can say that equals 2 x from 1 to n

realistically I think he could have just written only 1 of those 2nCn in the first place

scaryshark09 · Oct 19, 2023

Wait no now I’m confused

scaryshark09 · Oct 19, 2023

@Drongoski why did u have two of 2nCn in the first place?

Drongoski · Oct 19, 2023

scaryshark09 said:
@Drongoski why did u have two of 2nCn in the first place?

(1 + 1)^2n has 2n+1 (odd number!) of terms. The expression 2nC0 + 2nC1 + . . . + 2nCn has n+1 terms. By adding an additional 2nCn to the remaining expression, I make it equal to the 1st half. I did not notice this at the beginning and that's why I thought the identity was incorrect at first.

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