Prove that ${2n \choose n}= 2{2n-1 \choose n}$

Question

So I'm completely new to this, and I have a very basic understanding of how this works. Here is my best attempt at trying to prove this.

${2n \choose n}$

=$\frac{(2n)!}{(n)!(2n-n)!}$

=$\frac{(2n)!}{2(n)!}$

and then

${2n-1 \choose n}$

=$\frac{(2n-1)!}{(n)! (2n-1-n)!}$

=$\frac{(2n-1)!}{(n)! (n-1)!}$

I have no idea if this is the method to prove this problem, or even if I have done it correctly up to this point. I hit a block in trying to get the ${2n \choose n}$ equivalent to the ${2n-1 \choose n}$.

I am really at a beginner level here, so if you could explain your steps it would be great!

Answer 1

First of all you made a mistake when you wrote that $(n)!(n)!=2(n)!$. It should be $(n!)^2$. So $\binom{2n}{n}=\frac{(2n)!}{(n!)^2}$.

Next, you got that $\binom{2n-1}{n}=\frac{(2n-1)!}{(n)!(n-1)!}$. Now just multiply and divide the expression by $2n$. You will get $\frac{(2n)!}{2(n!)^2}$. Hence $2\binom{2n-1}{n}=\frac{(2n)!}{(n!)^2}=\binom{2n}{n}$.

Answer 2

Noting $2n-n=n, $ we have $${2n \choose n}=\frac{\color{red}{(2n)!}}{\color{green}{(n)!}(n)!}.$$

Also, noting $(2n-1)-n=n-1, $ we have $${2n-1 \choose n}=\frac{(2n-1)!}{(n)!(n-1)!}.$$

Finally, note that $\color{red}{(2n)!}=2n\times(2n-1)!$ and $\color{green}{n!}=n\times(n-1)!$ and $\frac {2n}n=2$ and we're done.

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The last note is because $m! = m\times(m-1)\times(m-2)\times...\times2\times1$ and

$(m-1)!=(m-1)\times (m-2)\times...\times2\times1$,

so $m!=m\times(m-1)!$.

Answer 3

With very few calculations:

We can use the recurrence relation $$\binom nk=\frac nk\binom{n-1}{k-1}, $$ which is the basis of the proof of the formula with factorials. In particular, $$\binom{2n}n= 2\binom{2n-1}{n-1}=2\binom{2n-1}n,\quad\text {since}\quad n=(2n-1)- (n-1).$$

Answer 4

$$\text{We know from Pascal's triangle that }\binom m k=\binom{m−1}{k−1}+\binom{m−1}{k}.$$ $$\therefore\binom{2n}n=\binom{2n−1}{n−1}+\binom{2n−1}n=2\binom{2n−1}n$$

$$\text{since }\binom{2n-1}{n-1}=\binom{2n-1}n\text{ since } (2n-1)-(n-1)=n.$$