show that \(\large \binom{2n}{n}\) is divisible by \(2\) for all \(n\ge 1\)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

show that \(\large \binom{2n}{n}\) is divisible by \(2\) for all \(n\ge 1\)

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\left(\begin{matrix}2n \\ n\end{matrix}\right) = \frac{ (2n)! }{ n!n! }\]
nvm
|dw:1436166061011:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Hmmm, I guess:\[ {2n\choose n} = {2n-1\choose n-1} + {2n-1 \choose n} = 2 {2n-1 \choose n} \]
wow! pascal's rule does the job is it
Lol that's actually pretty clever
I got excited about this problem because it has many solutions but what wio has is the most clever+short+simple one i think
So is that also skew symmetry
No, it isn't
whats skew symmetry ? i do see the left and right ends are symmetrical |dw:1436166600905:dw|
You could say: \[ f(k,m) ={k+m\choose k} = {m+k \choose m} = f(m,k) \]A function defined this way would be symmetric, kinda
It's not skew symmetry, but I do see symmetry obviously, skew symmetry is for a matrix (I actually learnt that yesterday from empty) haha...thought I could apply it at other places as well
\[aij = −aji\] xD
Skew symmetry is more like subtraction, I suppose: \[ f(x,y) = x-y = -(y-x) = -f(y,x) \]
Yeah you're right
But it's better to say 'commutative' than 'symmetric' when talking about maps, and to say 'symmetric' when talking about relations.
\[ f(k,m) ={k+m\choose k} = {m+k \choose m} = f(m,k) \] Again this follows from identity right \[\binom{n}{r} = \binom{n}{n-r}\]
thats an interesting way to see it
|dw:1436167013449:dw|
another proof which is just algebra but looks kinda cool \[x+\binom{2n}{n}+x = 2^{2n} \implies \binom{2n}{n} = 2(-x+2^{2n-1})\]
I like that one lol
\[ x = \sum_{k=0}^{n-1}{2n\choose k} = \sum_{k=0}^{n-1}{2n\choose 2n-k} \]
The odd rows have a nice closed form for the sum http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28%282n%2B1%29+choose+k%29
\[ \sum_{k=0}^{n}{2n+1\choose k} = \sum_{k=0}^{2n}{2n\choose k} \] not sure if this is an obvious result

Not the answer you are looking for?

Search for more explanations.

Ask your own question