What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$?












3












$begingroup$


What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$? That is, find $g(n)$ such that $$lim_{nto infty}frac{sum_{k=0}^n {nchoose k}^2}{g(n)}=1$$



We can expand the binomial coefficient and use Stirling's approximation but I can not determine g(n).






probability asymptotics







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    Were you able to simplify the summation to $binom{2n}{n}$?
    $endgroup$
    – JimmyK4542
    Dec 5 '18 at 2:27










  • $begingroup$
    From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 2:28










  • $begingroup$
    I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:37






  • 1




    $begingroup$
    What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:44










  • $begingroup$
    See here math.stackexchange.com/questions/74651/…
    $endgroup$
    – leonbloy
    Dec 5 '18 at 3:27
















3












$begingroup$


What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$? That is, find $g(n)$ such that $$lim_{nto infty}frac{sum_{k=0}^n {nchoose k}^2}{g(n)}=1$$



We can expand the binomial coefficient and use Stirling's approximation but I can not determine g(n).






probability asymptotics







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    Were you able to simplify the summation to $binom{2n}{n}$?
    $endgroup$
    – JimmyK4542
    Dec 5 '18 at 2:27










  • $begingroup$
    From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 2:28










  • $begingroup$
    I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:37






  • 1




    $begingroup$
    What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:44










  • $begingroup$
    See here math.stackexchange.com/questions/74651/…
    $endgroup$
    – leonbloy
    Dec 5 '18 at 3:27














3












3








3


1



$begingroup$


What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$? That is, find $g(n)$ such that $$lim_{nto infty}frac{sum_{k=0}^n {nchoose k}^2}{g(n)}=1$$



We can expand the binomial coefficient and use Stirling's approximation but I can not determine g(n).






probability asymptotics







share|cite|improve this question









$endgroup$




What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$? That is, find $g(n)$ such that $$lim_{nto infty}frac{sum_{k=0}^n {nchoose k}^2}{g(n)}=1$$



We can expand the binomial coefficient and use Stirling's approximation but I can not determine g(n).






probability asymptotics




probability asymptotics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 2:17









Probability studentProbability student

213




213








  • 6




    $begingroup$
    Were you able to simplify the summation to $binom{2n}{n}$?
    $endgroup$
    – JimmyK4542
    Dec 5 '18 at 2:27










  • $begingroup$
    From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 2:28










  • $begingroup$
    I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:37






  • 1




    $begingroup$
    What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:44










  • $begingroup$
    See here math.stackexchange.com/questions/74651/…
    $endgroup$
    – leonbloy
    Dec 5 '18 at 3:27














  • 6




    $begingroup$
    Were you able to simplify the summation to $binom{2n}{n}$?
    $endgroup$
    – JimmyK4542
    Dec 5 '18 at 2:27










  • $begingroup$
    From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 2:28










  • $begingroup$
    I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:37






  • 1




    $begingroup$
    What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:44










  • $begingroup$
    See here math.stackexchange.com/questions/74651/…
    $endgroup$
    – leonbloy
    Dec 5 '18 at 3:27








6




6




$begingroup$
Were you able to simplify the summation to $binom{2n}{n}$?
$endgroup$
– JimmyK4542
Dec 5 '18 at 2:27




$begingroup$
Were you able to simplify the summation to $binom{2n}{n}$?
$endgroup$
– JimmyK4542
Dec 5 '18 at 2:27












$begingroup$
From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
$endgroup$
– Henry Lee
Dec 5 '18 at 2:28




$begingroup$
From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
$endgroup$
– Henry Lee
Dec 5 '18 at 2:28












$begingroup$
I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
$endgroup$
– Probability student
Dec 5 '18 at 2:37




$begingroup$
I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
$endgroup$
– Probability student
Dec 5 '18 at 2:37




1




1




$begingroup$
What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
$endgroup$
– Probability student
Dec 5 '18 at 2:44




$begingroup$
What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
$endgroup$
– Probability student
Dec 5 '18 at 2:44












$begingroup$
See here math.stackexchange.com/questions/74651/…
$endgroup$
– leonbloy
Dec 5 '18 at 3:27




$begingroup$
See here math.stackexchange.com/questions/74651/…
$endgroup$
– leonbloy
Dec 5 '18 at 3:27










3 Answers
3






active

oldest

votes


















1












$begingroup$

$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47



















0












$begingroup$

This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09



















0












$begingroup$

If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026509%2fwhat-is-the-asymptotic-order-of-sum-k-0n-n-choose-k2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47
















1












$begingroup$

$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47














1












1








1





$begingroup$

$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.






share|cite|improve this answer











$endgroup$



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 3:47

























answered Dec 5 '18 at 3:14









Carl SchildkrautCarl Schildkraut

11.3k11441




11.3k11441












  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47


















  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47
















$begingroup$
Thanks Carl. I think it should be $${2n+m choose n}$$
$endgroup$
– Probability student
Dec 5 '18 at 3:19




$begingroup$
Thanks Carl. I think it should be $${2n+m choose n}$$
$endgroup$
– Probability student
Dec 5 '18 at 3:19












$begingroup$
@Probabilitystudent Thanks -- I've fixed it.
$endgroup$
– Carl Schildkraut
Dec 5 '18 at 3:47




$begingroup$
@Probabilitystudent Thanks -- I've fixed it.
$endgroup$
– Carl Schildkraut
Dec 5 '18 at 3:47











0












$begingroup$

This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09
















0












$begingroup$

This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09














0












0








0





$begingroup$

This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.






share|cite|improve this answer









$endgroup$



This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 3:00









marty cohenmarty cohen

73.3k549128




73.3k549128








  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09














  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09








1




1




$begingroup$
thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
$endgroup$
– Probability student
Dec 5 '18 at 3:04




$begingroup$
thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
$endgroup$
– Probability student
Dec 5 '18 at 3:04












$begingroup$
"Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
$endgroup$
– marty cohen
Dec 5 '18 at 3:09




$begingroup$
"Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
$endgroup$
– marty cohen
Dec 5 '18 at 3:09











0












$begingroup$

If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24


















0












$begingroup$

If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24
















0












0








0





$begingroup$

If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.






share|cite|improve this answer









$endgroup$



If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 4:20









Claude LeiboviciClaude Leibovici

120k1157132




120k1157132












  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24




















  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24


















$begingroup$
Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
$endgroup$
– Probability student
Dec 5 '18 at 10:24






$begingroup$
Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
$endgroup$
– Probability student
Dec 5 '18 at 10:24




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026509%2fwhat-is-the-asymptotic-order-of-sum-k-0n-n-choose-k2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more