Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Enumerative combinatorics, graph theory, order theory, posets, matroids, designs and other discrete structures. It also includes algebraic, analytic and probabilistic combinatorics.
11
votes
4
answers
1k
views
Combinatorial identity involving the square of $\binom{2n}{n}$
Is there any closed formula for
$$
\sum_{k=0}^n\frac{\binom{2k}{k}^2}{2^{4k}}
$$
?
This sum of is made out of the square of terms $a_{k}:=\frac{\binom{2k}{k}}{2^{2k}}$
I have been trying to verify …
3
votes
Combinatorial identity involving the square of $\binom{2n}{n}$
The limit I want to verify is
$$
\lim_{n\to\infty} (2n+1)\left[\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{\left(\sum_{j=0}^k a^2_{j}\right)}{(2k+1)(2k+2)}\right] -\frac{1}{2}{\sum_{k=0}^na^2_{k}}=\frac{1}{2 …