Linked Questions

7
votes
3answers
1k views

Proof of a combinatorial identity: $\sum_{i=0}^n {2i \choose i}{2(n-i)\choose n-i} = 4^n$ [duplicate]

Possible Duplicate: Identity involving binomial coefficients This was part of a homework assignment that I had, and I couldn't figure it out. Now it is bugging me. Can anyone help me? ...
4
votes
2answers
239 views

Finding a closed form expression for this sum [duplicate]

For non-negative $n$, find $$ \sum_{k=0}^n \binom{2k}{k}\binom{2n-2k}{n-k}. $$ I can't figure this out. Any ideas?
3
votes
3answers
179 views

Proving an Identity involving $4^N$ [duplicate]

I am trying to prove the following identity: $$\sum_{k=0}^N\left({2 \, N - 2 \, k \choose N - k}{2 \, k \choose k}\right)=4^N$$ I have tried writing $4^N=2^{2N}=(1+1)^{2N}=(1+1)^N(1+1)^N$, and ...
2
votes
1answer
170 views

How prove this $\sum_{k+j=n,0\le k,j\le n}\binom{2k}{k}\binom{2j}{j}=4^n$ [duplicate]

Show that $$ \sum_{k\ +\ j\ =\ n\atop{\vphantom{\LARGE A}0\ \le\ k,\phantom{A} j\ \le\ n}}{2k \choose k}{2j \choose j} = 4^{n} $$ I think use integral solve it. But I don't it,and this problem is ...
2
votes
0answers
107 views

How to prove $\sum_{k=1}^{n}\cot^2\left( \frac{k\pi}{2n+1}\right)=\frac{n(2n-1)}{3}$ [duplicate]

Possible Duplicate: Identity involving binomial coefficients how to prove $$ \sum_{k=1}^{n}\cot^2\left( \frac{k\pi}{2n+1}\right)=\frac{n(2n-1)}{3} $$ another $$ \sum_{k=0}^n{\binom {2k} ...
3
votes
0answers
95 views

Simplifying the sum with binomial coefficients [duplicate]

Possible Duplicate: Identity involving binomial coefficients Simplify the sum: $$\sum_{k=0}^n {2k\choose k}{2n-2k\choose n-k}$$ So we can denote $a_n=\sum_{k=0}^n {2k\choose ...
3
votes
0answers
38 views

Combinatorial proof of $\sum\limits_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} =4^n$ [duplicate]

$$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} = 4^n$$ Is there a combinatorial proof of above identity, without any arithmetic transformation? Thanks...
13
votes
5answers
538 views

Combinatorially showing $\lim_{n\to \infty}{\frac{2n\choose n}{4^n}}=0$

I am trying to show that $\lim_{n\to \infty}{\frac{2n\choose n}{4^n}}=0$. I found that using stirling's approximation, I can get: $$ \lim_{n\to \infty}{\frac{2n\choose n}{4^n}}= \lim_{n\to ...
47
votes
2answers
3k views

Combinatorial proof that $\sum \limits_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$ when $n$ is even

In my answer here I prove, using generating functions, a statement equivalent to $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$$ when $n$ is even. (Clearly the sum is ...
3
votes
3answers
3k views

Finding the power series of $\arcsin x$

I'm trying to find the power series of $\arcsin x$. This is what I did so far: $(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$, so $\arcsin x=\int \sqrt{\sum_{n=0}^{\infty}x^{2n}}$. (for $|x|<1$) Any ...
5
votes
2answers
921 views

Show $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$

How do you prove that $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$? I tried to identify the sum as a binomial series, but the $4$ and the $-1/2$ puzzle me. (This series arises in ...
8
votes
3answers
374 views

expected number of balls withdrawn to get equal numbers of black and white balls

There are $n$ black balls and $n$ white balls in a bin. I withdraw the balls one at a time without replacement until I have an equal number of white and black balls. What is the expected number of ...

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