Mixing
How is $left(frac{n!}{(2n +1)!!}right)^2 4^n = left(frac{(2n)!!}{(2n +1)!!}right)^2 ?$
1
$begingroup$
How is $$left(frac{n!}{(2n +1)!!}right)^2 4^n = left(frac{(2n)!!}{(2n +1)!!}right)^2 ?$$
Could anyone explain this for me please?
real-analysis calculus sequences-and-series analysis
edited Jan 29 at 10:34
Chinnapparaj R
5,8602928
asked Jan 29 at 10:27
hopefullyhopefully
301214
$endgroup$
add a comment |
1
$begingroup$
How is $$left(frac{n!}{(2n +1)!!}right)^2 4^n = left(frac{(2n)!!}{(2n +1)!!}right)^2 ?$$
Could anyone explain this for me please?
real-analysis calculus sequences-and-series analysis
edited Jan 29 at 10:34
Chinnapparaj R
5,8602928
asked Jan 29 at 10:27
hopefullyhopefully
301214
$endgroup$
$begingroup$
$2n!! =2ncdot (2n-2)cdot...=2^n(ncdot (n-1)cdot...)=2^nn! $
$endgroup$
– User123456789
Jan 29 at 10:32
$begingroup$
Your denominators are obviously equal - you need to show that $(n!)^24^n = ((2n)!!)^2$. Do you know what the double factorial notation represents?
$endgroup$
– Umberto P.
Jan 29 at 10:33
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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.
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– Carl Mummert
Jan 29 at 13:31
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Okay I will try to do this as much as I can ..... thank you for your advice @CarlMummert
$endgroup$
– hopefully
Jan 30 at 6:42
add a comment |
1
1
1
0
$begingroup$
How is $$left(frac{n!}{(2n +1)!!}right)^2 4^n = left(frac{(2n)!!}{(2n +1)!!}right)^2 ?$$
Could anyone explain this for me please?
real-analysis calculus sequences-and-series analysis
edited Jan 29 at 10:34
Chinnapparaj R
5,8602928
asked Jan 29 at 10:27
hopefullyhopefully
301214
$endgroup$
How is $$left(frac{n!}{(2n +1)!!}right)^2 4^n = left(frac{(2n)!!}{(2n +1)!!}right)^2 ?$$
Could anyone explain this for me please?
real-analysis calculus sequences-and-series analysis
real-analysis calculus sequences-and-series analysis
edited Jan 29 at 10:34
Chinnapparaj R
5,8602928
asked Jan 29 at 10:27
hopefullyhopefully
301214
edited Jan 29 at 10:34
Chinnapparaj R
5,8602928
asked Jan 29 at 10:27
hopefullyhopefully
301214
edited Jan 29 at 10:34
Chinnapparaj R
5,8602928
edited Jan 29 at 10:34
Chinnapparaj R
5,8602928
edited Jan 29 at 10:34
Chinnapparaj R
5,8602928
5,8602928
asked Jan 29 at 10:27
hopefullyhopefully
301214
asked Jan 29 at 10:27
hopefullyhopefully
301214
asked Jan 29 at 10:27
hopefullyhopefully
301214
301214
$begingroup$
$2n!! =2ncdot (2n-2)cdot...=2^n(ncdot (n-1)cdot...)=2^nn! $
$endgroup$
– User123456789
Jan 29 at 10:32
$begingroup$
Your denominators are obviously equal - you need to show that $(n!)^24^n = ((2n)!!)^2$. Do you know what the double factorial notation represents?
$endgroup$
– Umberto P.
Jan 29 at 10:33
$begingroup$
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
$begingroup$
Okay I will try to do this as much as I can ..... thank you for your advice @CarlMummert
$endgroup$
– hopefully
Jan 30 at 6:42
add a comment |
$begingroup$
$2n!! =2ncdot (2n-2)cdot...=2^n(ncdot (n-1)cdot...)=2^nn! $
$endgroup$
– User123456789
Jan 29 at 10:32
$begingroup$
Your denominators are obviously equal - you need to show that $(n!)^24^n = ((2n)!!)^2$. Do you know what the double factorial notation represents?
$endgroup$
– Umberto P.
Jan 29 at 10:33
$begingroup$
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
$begingroup$
Okay I will try to do this as much as I can ..... thank you for your advice @CarlMummert
$endgroup$
– hopefully
Jan 30 at 6:42
$begingroup$
$2n!! =2ncdot (2n-2)cdot...=2^n(ncdot (n-1)cdot...)=2^nn! $
$endgroup$
– User123456789
Jan 29 at 10:32
$begingroup$
$2n!! =2ncdot (2n-2)cdot...=2^n(ncdot (n-1)cdot...)=2^nn! $
$endgroup$
– User123456789
Jan 29 at 10:32
$begingroup$
Your denominators are obviously equal - you need to show that $(n!)^24^n = ((2n)!!)^2$. Do you know what the double factorial notation represents?
$endgroup$
– Umberto P.
Jan 29 at 10:33
$begingroup$
Your denominators are obviously equal - you need to show that $(n!)^24^n = ((2n)!!)^2$. Do you know what the double factorial notation represents?
$endgroup$
– Umberto P.
Jan 29 at 10:33
$begingroup$
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
$begingroup$
Okay I will try to do this as much as I can ..... thank you for your advice @CarlMummert
$endgroup$
– hopefully
Jan 30 at 6:42
$begingroup$
Okay I will try to do this as much as I can ..... thank you for your advice @CarlMummert
$endgroup$
– hopefully
Jan 30 at 6:42
add a comment |
1 Answer
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$begingroup$
Some hints:
Since the denominators of both expressions are the same, all you really need to prove is that $$ (n!)^24^n = ((2n)!!)^2 $$
$$begin{align}(2n)!! &= 2ncdot (2n-2)cdot (2n-4)cdots 2 \&= 2cdot ncdot2cdot(n-1)cdot2cdot(n-2)cdots2cdot 1 \&= 2cdot2cdots 2cdot ncdot(n-1)cdots 1end{align}$$
edited Jan 29 at 10:44
user549397
1,6541418
answered Jan 29 at 10:32
5xum5xum
91.9k394161
$endgroup$
add a comment |
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1 Answer
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1 Answer
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2
$begingroup$
Some hints:
Since the denominators of both expressions are the same, all you really need to prove is that $$ (n!)^24^n = ((2n)!!)^2 $$
$$begin{align}(2n)!! &= 2ncdot (2n-2)cdot (2n-4)cdots 2 \&= 2cdot ncdot2cdot(n-1)cdot2cdot(n-2)cdots2cdot 1 \&= 2cdot2cdots 2cdot ncdot(n-1)cdots 1end{align}$$
edited Jan 29 at 10:44
user549397
1,6541418
answered Jan 29 at 10:32
5xum5xum
91.9k394161
$endgroup$
add a comment |
2
$begingroup$
Some hints:
Since the denominators of both expressions are the same, all you really need to prove is that $$ (n!)^24^n = ((2n)!!)^2 $$
$$begin{align}(2n)!! &= 2ncdot (2n-2)cdot (2n-4)cdots 2 \&= 2cdot ncdot2cdot(n-1)cdot2cdot(n-2)cdots2cdot 1 \&= 2cdot2cdots 2cdot ncdot(n-1)cdots 1end{align}$$
edited Jan 29 at 10:44
user549397
1,6541418
answered Jan 29 at 10:32
5xum5xum
91.9k394161
$endgroup$
add a comment |
2
2
2
$begingroup$
Some hints:
Since the denominators of both expressions are the same, all you really need to prove is that $$ (n!)^24^n = ((2n)!!)^2 $$
$$begin{align}(2n)!! &= 2ncdot (2n-2)cdot (2n-4)cdots 2 \&= 2cdot ncdot2cdot(n-1)cdot2cdot(n-2)cdots2cdot 1 \&= 2cdot2cdots 2cdot ncdot(n-1)cdots 1end{align}$$
edited Jan 29 at 10:44
user549397
1,6541418
answered Jan 29 at 10:32
5xum5xum
91.9k394161
$endgroup$
Some hints:
Since the denominators of both expressions are the same, all you really need to prove is that $$ (n!)^24^n = ((2n)!!)^2 $$
$$begin{align}(2n)!! &= 2ncdot (2n-2)cdot (2n-4)cdots 2 \&= 2cdot ncdot2cdot(n-1)cdot2cdot(n-2)cdots2cdot 1 \&= 2cdot2cdots 2cdot ncdot(n-1)cdots 1end{align}$$
edited Jan 29 at 10:44
user549397
1,6541418
answered Jan 29 at 10:32
5xum5xum
91.9k394161
edited Jan 29 at 10:44
user549397
1,6541418
edited Jan 29 at 10:44
user549397
1,6541418
edited Jan 29 at 10:44
user549397
1,6541418
1,6541418
answered Jan 29 at 10:32
5xum5xum
91.9k394161
answered Jan 29 at 10:32
5xum5xum
91.9k394161
answered Jan 29 at 10:32
5xum5xum
91.9k394161
91.9k394161
add a comment |
add a comment |
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$begingroup$
$2n!! =2ncdot (2n-2)cdot...=2^n(ncdot (n-1)cdot...)=2^nn! $
$endgroup$
– User123456789
Jan 29 at 10:32
$begingroup$
Your denominators are obviously equal - you need to show that $(n!)^24^n = ((2n)!!)^2$. Do you know what the double factorial notation represents?
$endgroup$
– Umberto P.
Jan 29 at 10:33
$begingroup$
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
$begingroup$
Okay I will try to do this as much as I can ..... thank you for your advice @CarlMummert
$endgroup$
– hopefully
Jan 30 at 6:42