Why is $prod_{k=n}^{2n-3}left(2n-kright) = prod_{j=3}^{n}j$?
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I am trying to figure out these two equal expressions from my textbook:
$$prod_{k=n}^{2n-3}left(2n-kright) = prod _ {j=3} ^ {n}j$$
I have checked them and know that they apply, but what are the logical steps that lead me to the equation?
sequences-and-series discrete-mathematics
asked Dec 19 '18 at 17:00
LuckyLucky
1127
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add a comment |
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I am trying to figure out these two equal expressions from my textbook:
$$prod_{k=n}^{2n-3}left(2n-kright) = prod _ {j=3} ^ {n}j$$
I have checked them and know that they apply, but what are the logical steps that lead me to the equation?
sequences-and-series discrete-mathematics
asked Dec 19 '18 at 17:00
LuckyLucky
1127
$endgroup$
add a comment |
$begingroup$
I am trying to figure out these two equal expressions from my textbook:
$$prod_{k=n}^{2n-3}left(2n-kright) = prod _ {j=3} ^ {n}j$$
I have checked them and know that they apply, but what are the logical steps that lead me to the equation?
sequences-and-series discrete-mathematics
asked Dec 19 '18 at 17:00
LuckyLucky
1127
$endgroup$
I am trying to figure out these two equal expressions from my textbook:
$$prod_{k=n}^{2n-3}left(2n-kright) = prod _ {j=3} ^ {n}j$$
I have checked them and know that they apply, but what are the logical steps that lead me to the equation?
sequences-and-series discrete-mathematics
sequences-and-series discrete-mathematics
asked Dec 19 '18 at 17:00
LuckyLucky
1127
asked Dec 19 '18 at 17:00
LuckyLucky
1127
edited Dec 19 '18 at 17:13
Lucky
asked Dec 19 '18 at 17:00
LuckyLucky
1127
asked Dec 19 '18 at 17:00
LuckyLucky
1127
asked Dec 19 '18 at 17:00
LuckyLucky
1127
1127
add a comment |
add a comment |
3 Answers
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Let's write out the terms and see if we can observe a pattern.
$$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
Which simplifies to
$$n(n-1)(n-2)(n-3)...(3)$$
Which is equal to the desired product, since multiplication is commutative.
answered Dec 19 '18 at 17:07
user667user667
164117
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Hint: Using a simple change variable:
$$j = 2n - k$$
Hence, its range is:
$$ 2n - n = n geq j geq 2n-(2n-3) = 3$$
answered Dec 19 '18 at 17:04
OmGOmG
2,502822
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begin{equation}
prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
end{equation}
Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$
answered Dec 19 '18 at 17:14
Ahmad BazziAhmad Bazzi
8,2962824
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3 Answers
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3 Answers
3
active
oldest
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$begingroup$
Let's write out the terms and see if we can observe a pattern.
$$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
Which simplifies to
$$n(n-1)(n-2)(n-3)...(3)$$
Which is equal to the desired product, since multiplication is commutative.
answered Dec 19 '18 at 17:07
user667user667
164117
$endgroup$
add a comment |
$begingroup$
Let's write out the terms and see if we can observe a pattern.
$$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
Which simplifies to
$$n(n-1)(n-2)(n-3)...(3)$$
Which is equal to the desired product, since multiplication is commutative.
answered Dec 19 '18 at 17:07
user667user667
164117
$endgroup$
add a comment |
$begingroup$
Let's write out the terms and see if we can observe a pattern.
$$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
Which simplifies to
$$n(n-1)(n-2)(n-3)...(3)$$
Which is equal to the desired product, since multiplication is commutative.
answered Dec 19 '18 at 17:07
user667user667
164117
$endgroup$
Let's write out the terms and see if we can observe a pattern.
$$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
Which simplifies to
$$n(n-1)(n-2)(n-3)...(3)$$
Which is equal to the desired product, since multiplication is commutative.
answered Dec 19 '18 at 17:07
user667user667
164117
answered Dec 19 '18 at 17:07
user667user667
164117
answered Dec 19 '18 at 17:07
user667user667
164117
answered Dec 19 '18 at 17:07
user667user667
164117
164117
add a comment |
add a comment |
$begingroup$
Hint: Using a simple change variable:
$$j = 2n - k$$
Hence, its range is:
$$ 2n - n = n geq j geq 2n-(2n-3) = 3$$
answered Dec 19 '18 at 17:04
OmGOmG
2,502822
$endgroup$
add a comment |
$begingroup$
Hint: Using a simple change variable:
$$j = 2n - k$$
Hence, its range is:
$$ 2n - n = n geq j geq 2n-(2n-3) = 3$$
answered Dec 19 '18 at 17:04
OmGOmG
2,502822
$endgroup$
add a comment |
$begingroup$
Hint: Using a simple change variable:
$$j = 2n - k$$
Hence, its range is:
$$ 2n - n = n geq j geq 2n-(2n-3) = 3$$
answered Dec 19 '18 at 17:04
OmGOmG
2,502822
$endgroup$
Hint: Using a simple change variable:
$$j = 2n - k$$
Hence, its range is:
$$ 2n - n = n geq j geq 2n-(2n-3) = 3$$
answered Dec 19 '18 at 17:04
OmGOmG
2,502822
answered Dec 19 '18 at 17:04
OmGOmG
2,502822
answered Dec 19 '18 at 17:04
OmGOmG
2,502822
answered Dec 19 '18 at 17:04
OmGOmG
2,502822
2,502822
add a comment |
add a comment |
$begingroup$
begin{equation}
prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
end{equation}
Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$
answered Dec 19 '18 at 17:14
Ahmad BazziAhmad Bazzi
8,2962824
$endgroup$
add a comment |
$begingroup$
begin{equation}
prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
end{equation}
Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$
answered Dec 19 '18 at 17:14
Ahmad BazziAhmad Bazzi
8,2962824
$endgroup$
add a comment |
$begingroup$
begin{equation}
prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
end{equation}
Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$
answered Dec 19 '18 at 17:14
Ahmad BazziAhmad Bazzi
8,2962824
$endgroup$
begin{equation}
prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
end{equation}
Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$
answered Dec 19 '18 at 17:14
Ahmad BazziAhmad Bazzi
8,2962824
answered Dec 19 '18 at 17:14
Ahmad BazziAhmad Bazzi
8,2962824
answered Dec 19 '18 at 17:14
Ahmad BazziAhmad Bazzi
8,2962824
answered Dec 19 '18 at 17:14
Ahmad BazziAhmad Bazzi
8,2962824
8,2962824
add a comment |
add a comment |
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