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







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  • 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











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share|cite|improve this question










asked Jan 12 at 7:18









LuciferLucifer

453




453








  • 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
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$begingroup$

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






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  • 3




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












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1 Answer
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1 Answer
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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

4,7382727




4,7382727








  • 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


















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