A query in binomial th. Based question












3












$begingroup$


Posting the part of the solution...i know the concept on which it is asked but i got stuck while solving and i peeked into the solution, i found this



$=frac{2n(2n-1)(2n-2)...4.3.2.1}{n!n!}x^n$ fine
$=frac{1.2.3.4...(2n-2)(2n-1)(2n)}{n!n!}x^n$ fine
$=frac{[1.3.5...(2n-1)][2.4.6...(2n)]}{n!n!}x^n$ fine
$=frac{[1.3.5..(2n-1)]2^n[1.2.3...n]}{n!n!}x^n$ 2^n is taken as common, i'm tricked here, n is raised to 2 which means there are n terms into $“[2.4.6...(2n)]”$ but how?
Please explain how there are n terms into that expression ($“[2.4.6...(2n)]”$)






binomial-theorem







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is not clear what you're asking; there is an equals sign with nothing before it.
    $endgroup$
    – Carl Mummert
    Jan 29 at 12:19






  • 1




    $begingroup$
    While StackExchange is a Q/A site, more than just a question is expected from the asker. Try to focus your question to an actual and specific problem you have faced. If you can, include details about what you have tried and exactly what you are trying to do. If you are stuck, provide definitions and your own background. Write down what you know and generally try to be constructive. If you can, provide motivation and context for the problem. This is meant to be an exchange, both ways, as the site name implies.
    $endgroup$
    – Carl Mummert
    Jan 29 at 13:31
















3












$begingroup$


Posting the part of the solution...i know the concept on which it is asked but i got stuck while solving and i peeked into the solution, i found this



$=frac{2n(2n-1)(2n-2)...4.3.2.1}{n!n!}x^n$ fine
$=frac{1.2.3.4...(2n-2)(2n-1)(2n)}{n!n!}x^n$ fine
$=frac{[1.3.5...(2n-1)][2.4.6...(2n)]}{n!n!}x^n$ fine
$=frac{[1.3.5..(2n-1)]2^n[1.2.3...n]}{n!n!}x^n$ 2^n is taken as common, i'm tricked here, n is raised to 2 which means there are n terms into $“[2.4.6...(2n)]”$ but how?
Please explain how there are n terms into that expression ($“[2.4.6...(2n)]”$)






binomial-theorem







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is not clear what you're asking; there is an equals sign with nothing before it.
    $endgroup$
    – Carl Mummert
    Jan 29 at 12:19






  • 1




    $begingroup$
    While StackExchange is a Q/A site, more than just a question is expected from the asker. Try to focus your question to an actual and specific problem you have faced. If you can, include details about what you have tried and exactly what you are trying to do. If you are stuck, provide definitions and your own background. Write down what you know and generally try to be constructive. If you can, provide motivation and context for the problem. This is meant to be an exchange, both ways, as the site name implies.
    $endgroup$
    – Carl Mummert
    Jan 29 at 13:31














3












3








3





$begingroup$


Posting the part of the solution...i know the concept on which it is asked but i got stuck while solving and i peeked into the solution, i found this



$=frac{2n(2n-1)(2n-2)...4.3.2.1}{n!n!}x^n$ fine
$=frac{1.2.3.4...(2n-2)(2n-1)(2n)}{n!n!}x^n$ fine
$=frac{[1.3.5...(2n-1)][2.4.6...(2n)]}{n!n!}x^n$ fine
$=frac{[1.3.5..(2n-1)]2^n[1.2.3...n]}{n!n!}x^n$ 2^n is taken as common, i'm tricked here, n is raised to 2 which means there are n terms into $“[2.4.6...(2n)]”$ but how?
Please explain how there are n terms into that expression ($“[2.4.6...(2n)]”$)






binomial-theorem







share|cite|improve this question









$endgroup$




Posting the part of the solution...i know the concept on which it is asked but i got stuck while solving and i peeked into the solution, i found this



$=frac{2n(2n-1)(2n-2)...4.3.2.1}{n!n!}x^n$ fine
$=frac{1.2.3.4...(2n-2)(2n-1)(2n)}{n!n!}x^n$ fine
$=frac{[1.3.5...(2n-1)][2.4.6...(2n)]}{n!n!}x^n$ fine
$=frac{[1.3.5..(2n-1)]2^n[1.2.3...n]}{n!n!}x^n$ 2^n is taken as common, i'm tricked here, n is raised to 2 which means there are n terms into $“[2.4.6...(2n)]”$ but how?
Please explain how there are n terms into that expression ($“[2.4.6...(2n)]”$)






binomial-theorem




binomial-theorem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 7:18









LuciferLucifer

303




303








  • 1




    $begingroup$
    It is not clear what you're asking; there is an equals sign with nothing before it.
    $endgroup$
    – Carl Mummert
    Jan 29 at 12:19






  • 1




    $begingroup$
    While StackExchange is a Q/A site, more than just a question is expected from the asker. Try to focus your question to an actual and specific problem you have faced. If you can, include details about what you have tried and exactly what you are trying to do. If you are stuck, provide definitions and your own background. Write down what you know and generally try to be constructive. If you can, provide motivation and context for the problem. This is meant to be an exchange, both ways, as the site name implies.
    $endgroup$
    – Carl Mummert
    Jan 29 at 13:31














  • 1




    $begingroup$
    It is not clear what you're asking; there is an equals sign with nothing before it.
    $endgroup$
    – Carl Mummert
    Jan 29 at 12:19






  • 1




    $begingroup$
    While StackExchange is a Q/A site, more than just a question is expected from the asker. Try to focus your question to an actual and specific problem you have faced. If you can, include details about what you have tried and exactly what you are trying to do. If you are stuck, provide definitions and your own background. Write down what you know and generally try to be constructive. If you can, provide motivation and context for the problem. This is meant to be an exchange, both ways, as the site name implies.
    $endgroup$
    – Carl Mummert
    Jan 29 at 13:31








1




1




$begingroup$
It is not clear what you're asking; there is an equals sign with nothing before it.
$endgroup$
– Carl Mummert
Jan 29 at 12:19




$begingroup$
It is not clear what you're asking; there is an equals sign with nothing before it.
$endgroup$
– Carl Mummert
Jan 29 at 12:19




1




1




$begingroup$
While StackExchange is a Q/A site, more than just a question is expected from the asker. Try to focus your question to an actual and specific problem you have faced. If you can, include details about what you have tried and exactly what you are trying to do. If you are stuck, provide definitions and your own background. Write down what you know and generally try to be constructive. If you can, provide motivation and context for the problem. This is meant to be an exchange, both ways, as the site name implies.
$endgroup$
– Carl Mummert
Jan 29 at 13:31




$begingroup$
While StackExchange is a Q/A site, more than just a question is expected from the asker. Try to focus your question to an actual and specific problem you have faced. If you can, include details about what you have tried and exactly what you are trying to do. If you are stuck, provide definitions and your own background. Write down what you know and generally try to be constructive. If you can, provide motivation and context for the problem. This is meant to be an exchange, both ways, as the site name implies.
$endgroup$
– Carl Mummert
Jan 29 at 13:31










1 Answer
1






active

oldest

votes


















2












$begingroup$

$begin{align}
2×4×cdots ×(2n)&=(2×1)×(2×2)×(2×3)×dots×(2×n)\
&=2^n×(1×2×cdots×n )
end{align}$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Thank you .....i got it
    $endgroup$
    – Lucifer
    Jan 12 at 7:28











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%2f3070667%2fa-query-in-binomial-th-based-question%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$begin{align}
2×4×cdots ×(2n)&=(2×1)×(2×2)×(2×3)×dots×(2×n)\
&=2^n×(1×2×cdots×n )
end{align}$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Thank you .....i got it
    $endgroup$
    – Lucifer
    Jan 12 at 7:28
















2












$begingroup$

$begin{align}
2×4×cdots ×(2n)&=(2×1)×(2×2)×(2×3)×dots×(2×n)\
&=2^n×(1×2×cdots×n )
end{align}$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Thank you .....i got it
    $endgroup$
    – Lucifer
    Jan 12 at 7:28














2












2








2





$begingroup$

$begin{align}
2×4×cdots ×(2n)&=(2×1)×(2×2)×(2×3)×dots×(2×n)\
&=2^n×(1×2×cdots×n )
end{align}$






share|cite|improve this answer









$endgroup$



$begin{align}
2×4×cdots ×(2n)&=(2×1)×(2×2)×(2×3)×dots×(2×n)\
&=2^n×(1×2×cdots×n )
end{align}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 7:25









Thomas ShelbyThomas Shelby

2,9971422




2,9971422








  • 3




    $begingroup$
    Thank you .....i got it
    $endgroup$
    – Lucifer
    Jan 12 at 7:28














  • 3




    $begingroup$
    Thank you .....i got it
    $endgroup$
    – Lucifer
    Jan 12 at 7:28








3




3




$begingroup$
Thank you .....i got it
$endgroup$
– Lucifer
Jan 12 at 7:28




$begingroup$
Thank you .....i got it
$endgroup$
– Lucifer
Jan 12 at 7:28


















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%2f3070667%2fa-query-in-binomial-th-based-question%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

Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

Image
28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
Read more

Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

Image
up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
Read more

A Topological Invariant for $pi_3(U(n))$

Image
3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
Read more