Mixing
Binomial coefficient from a double sequence ${a_{n,k}}$ of Pascal Triangle
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I really need help with the 13-th prob. Chap. 2 in the book Kuratowski "Introduction to Calculus", it says:
We consider the following table (i.e. double sequence ${a_{n,k}}$):
begin{array}{cccccc}
1 & 1 & 1 & 1 & 1 & cdots \
1 & 2 & 3 & 4 & 5 & cdots \
1 & 3 & 6 & 10 & 15 & cdots \
1 & 4 & 10 & 20 & 35 & cdots \
1 & 5 & 15 & 35 & 70 & cdots \
cdots & cdots & cdots & cdots &cdots & cdots \
end{array}
This table is defined as follows: the firs line consists of nothing but 1, i.e. $a_{1,k}=1$; the $k$-th term in the $n$-th line is the sum of the first $k$ terms of the $(n-1)$-th line, i.e.
begin{equation}
a_{n,k}=a_{n-1,1}+a_{n-1,2}+ cdots + a_{n-1,k}.
end{equation}
Prove that $$a_{n,k}=frac{(k+n-2)!}{(k-1)!(n-1)!}$$ and that the terms $a_{n,1}$, $a_{n-1,1}$,..., $a_{1,n}$ are successive coefficients of the Newton expansion of the expression $(a+b)^{n-1}$ (as is easily seen they are the terms of the table lying on the straight line joining the $n$-th term of the first line with the first term of the $n$-th line).
I tried to do induction but i don't know what to do with the two indexes, and when i tried to do something with the formula $a_{n,k}$ it becomes to difficult to manipulate.
I'll appreciate all the help or ideas.
calculus sequences-and-series number-theory binomial-coefficients
edited 2 days ago
darij grinberg
9,89532961
asked 2 days ago
RichardMoon15
53
add a comment |
up vote
0
down vote
favorite
I really need help with the 13-th prob. Chap. 2 in the book Kuratowski "Introduction to Calculus", it says:
We consider the following table (i.e. double sequence ${a_{n,k}}$):
begin{array}{cccccc}
1 & 1 & 1 & 1 & 1 & cdots \
1 & 2 & 3 & 4 & 5 & cdots \
1 & 3 & 6 & 10 & 15 & cdots \
1 & 4 & 10 & 20 & 35 & cdots \
1 & 5 & 15 & 35 & 70 & cdots \
cdots & cdots & cdots & cdots &cdots & cdots \
end{array}
This table is defined as follows: the firs line consists of nothing but 1, i.e. $a_{1,k}=1$; the $k$-th term in the $n$-th line is the sum of the first $k$ terms of the $(n-1)$-th line, i.e.
begin{equation}
a_{n,k}=a_{n-1,1}+a_{n-1,2}+ cdots + a_{n-1,k}.
end{equation}
Prove that $$a_{n,k}=frac{(k+n-2)!}{(k-1)!(n-1)!}$$ and that the terms $a_{n,1}$, $a_{n-1,1}$,..., $a_{1,n}$ are successive coefficients of the Newton expansion of the expression $(a+b)^{n-1}$ (as is easily seen they are the terms of the table lying on the straight line joining the $n$-th term of the first line with the first term of the $n$-th line).
I tried to do induction but i don't know what to do with the two indexes, and when i tried to do something with the formula $a_{n,k}$ it becomes to difficult to manipulate.
I'll appreciate all the help or ideas.
calculus sequences-and-series number-theory binomial-coefficients
edited 2 days ago
darij grinberg
9,89532961
asked 2 days ago
RichardMoon15
53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I really need help with the 13-th prob. Chap. 2 in the book Kuratowski "Introduction to Calculus", it says:
We consider the following table (i.e. double sequence ${a_{n,k}}$):
begin{array}{cccccc}
1 & 1 & 1 & 1 & 1 & cdots \
1 & 2 & 3 & 4 & 5 & cdots \
1 & 3 & 6 & 10 & 15 & cdots \
1 & 4 & 10 & 20 & 35 & cdots \
1 & 5 & 15 & 35 & 70 & cdots \
cdots & cdots & cdots & cdots &cdots & cdots \
end{array}
This table is defined as follows: the firs line consists of nothing but 1, i.e. $a_{1,k}=1$; the $k$-th term in the $n$-th line is the sum of the first $k$ terms of the $(n-1)$-th line, i.e.
begin{equation}
a_{n,k}=a_{n-1,1}+a_{n-1,2}+ cdots + a_{n-1,k}.
end{equation}
Prove that $$a_{n,k}=frac{(k+n-2)!}{(k-1)!(n-1)!}$$ and that the terms $a_{n,1}$, $a_{n-1,1}$,..., $a_{1,n}$ are successive coefficients of the Newton expansion of the expression $(a+b)^{n-1}$ (as is easily seen they are the terms of the table lying on the straight line joining the $n$-th term of the first line with the first term of the $n$-th line).
I tried to do induction but i don't know what to do with the two indexes, and when i tried to do something with the formula $a_{n,k}$ it becomes to difficult to manipulate.
I'll appreciate all the help or ideas.
calculus sequences-and-series number-theory binomial-coefficients
edited 2 days ago
darij grinberg
9,89532961
asked 2 days ago
RichardMoon15
53
I really need help with the 13-th prob. Chap. 2 in the book Kuratowski "Introduction to Calculus", it says:
We consider the following table (i.e. double sequence ${a_{n,k}}$):
begin{array}{cccccc}
1 & 1 & 1 & 1 & 1 & cdots \
1 & 2 & 3 & 4 & 5 & cdots \
1 & 3 & 6 & 10 & 15 & cdots \
1 & 4 & 10 & 20 & 35 & cdots \
1 & 5 & 15 & 35 & 70 & cdots \
cdots & cdots & cdots & cdots &cdots & cdots \
end{array}
This table is defined as follows: the firs line consists of nothing but 1, i.e. $a_{1,k}=1$; the $k$-th term in the $n$-th line is the sum of the first $k$ terms of the $(n-1)$-th line, i.e.
begin{equation}
a_{n,k}=a_{n-1,1}+a_{n-1,2}+ cdots + a_{n-1,k}.
end{equation}
Prove that $$a_{n,k}=frac{(k+n-2)!}{(k-1)!(n-1)!}$$ and that the terms $a_{n,1}$, $a_{n-1,1}$,..., $a_{1,n}$ are successive coefficients of the Newton expansion of the expression $(a+b)^{n-1}$ (as is easily seen they are the terms of the table lying on the straight line joining the $n$-th term of the first line with the first term of the $n$-th line).
I tried to do induction but i don't know what to do with the two indexes, and when i tried to do something with the formula $a_{n,k}$ it becomes to difficult to manipulate.
I'll appreciate all the help or ideas.
calculus sequences-and-series number-theory binomial-coefficients
calculus sequences-and-series number-theory binomial-coefficients
edited 2 days ago
darij grinberg
9,89532961
asked 2 days ago
RichardMoon15
53
edited 2 days ago
darij grinberg
9,89532961
asked 2 days ago
RichardMoon15
53
edited 2 days ago
darij grinberg
9,89532961
edited 2 days ago
darij grinberg
9,89532961
edited 2 days ago
darij grinberg
9,89532961
9,89532961
asked 2 days ago
RichardMoon15
53
asked 2 days ago
RichardMoon15
53
asked 2 days ago
RichardMoon15
53
53
add a comment |
add a comment |
1 Answer
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We can work by induction on the quantity $Y=k+n$. If $Y=2$ then $k=1$ and $n=1$ so
$$
a_{1, 1}=1=frac{(1+1-2)!}{(1-1)!(1-1)!}
$$
Now suppose that for a certain $Y$ the equality holds for every $k, n$ such that $Y=k+n$, so now set $n, k$ such that $n+k=Y+1$. If $n=1$ then
$$
a_{1, k}=frac{(k-1)!}{(k-1)!}=1
$$
Otherwise if $n>1$ we have
$$
a_{n, k}=a_{n-1, 1}+a_{n-1, 2}+dotsb+a_{n-1, k-1}+a_{n-1, k}=a_{n, k-1}+a_{n-1, k}
$$
with $n+k-1=n-1+k=Y$ and so we can use induction to obtain
$$
a_{n, k}=frac{(k+n-3)!}{(k-2)!(n-1)!}+frac{(k+n-3)!}{(k-1)!(n-2)!}=frac{(k+n-3)!}{(k-2)!(n-2)!(n-1)}+frac{(k+n-3)!}{(k-2)!(n-2)!(k-1)}=frac{(k+n-3)!(k-1)+(k+n-3)!(n-1)}{(k-2)!(n-2)!(k-1)(n-1)}=frac{(k+n-3)!(k+n-2)}{(k-1)!(n-1)!}=frac{(k+n-2)!}{(k-1)!(n-1)!}
$$
Obtaining so the initial statement.
answered 2 days ago
P De Donato
2236
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We can work by induction on the quantity $Y=k+n$. If $Y=2$ then $k=1$ and $n=1$ so
$$
a_{1, 1}=1=frac{(1+1-2)!}{(1-1)!(1-1)!}
$$
Now suppose that for a certain $Y$ the equality holds for every $k, n$ such that $Y=k+n$, so now set $n, k$ such that $n+k=Y+1$. If $n=1$ then
$$
a_{1, k}=frac{(k-1)!}{(k-1)!}=1
$$
Otherwise if $n>1$ we have
$$
a_{n, k}=a_{n-1, 1}+a_{n-1, 2}+dotsb+a_{n-1, k-1}+a_{n-1, k}=a_{n, k-1}+a_{n-1, k}
$$
with $n+k-1=n-1+k=Y$ and so we can use induction to obtain
$$
a_{n, k}=frac{(k+n-3)!}{(k-2)!(n-1)!}+frac{(k+n-3)!}{(k-1)!(n-2)!}=frac{(k+n-3)!}{(k-2)!(n-2)!(n-1)}+frac{(k+n-3)!}{(k-2)!(n-2)!(k-1)}=frac{(k+n-3)!(k-1)+(k+n-3)!(n-1)}{(k-2)!(n-2)!(k-1)(n-1)}=frac{(k+n-3)!(k+n-2)}{(k-1)!(n-1)!}=frac{(k+n-2)!}{(k-1)!(n-1)!}
$$
Obtaining so the initial statement.
answered 2 days ago
P De Donato
2236
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
0
down vote
We can work by induction on the quantity $Y=k+n$. If $Y=2$ then $k=1$ and $n=1$ so
$$
a_{1, 1}=1=frac{(1+1-2)!}{(1-1)!(1-1)!}
$$
Now suppose that for a certain $Y$ the equality holds for every $k, n$ such that $Y=k+n$, so now set $n, k$ such that $n+k=Y+1$. If $n=1$ then
$$
a_{1, k}=frac{(k-1)!}{(k-1)!}=1
$$
Otherwise if $n>1$ we have
$$
a_{n, k}=a_{n-1, 1}+a_{n-1, 2}+dotsb+a_{n-1, k-1}+a_{n-1, k}=a_{n, k-1}+a_{n-1, k}
$$
with $n+k-1=n-1+k=Y$ and so we can use induction to obtain
$$
a_{n, k}=frac{(k+n-3)!}{(k-2)!(n-1)!}+frac{(k+n-3)!}{(k-1)!(n-2)!}=frac{(k+n-3)!}{(k-2)!(n-2)!(n-1)}+frac{(k+n-3)!}{(k-2)!(n-2)!(k-1)}=frac{(k+n-3)!(k-1)+(k+n-3)!(n-1)}{(k-2)!(n-2)!(k-1)(n-1)}=frac{(k+n-3)!(k+n-2)}{(k-1)!(n-1)!}=frac{(k+n-2)!}{(k-1)!(n-1)!}
$$
Obtaining so the initial statement.
answered 2 days ago
P De Donato
2236
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
We can work by induction on the quantity $Y=k+n$. If $Y=2$ then $k=1$ and $n=1$ so
$$
a_{1, 1}=1=frac{(1+1-2)!}{(1-1)!(1-1)!}
$$
Now suppose that for a certain $Y$ the equality holds for every $k, n$ such that $Y=k+n$, so now set $n, k$ such that $n+k=Y+1$. If $n=1$ then
$$
a_{1, k}=frac{(k-1)!}{(k-1)!}=1
$$
Otherwise if $n>1$ we have
$$
a_{n, k}=a_{n-1, 1}+a_{n-1, 2}+dotsb+a_{n-1, k-1}+a_{n-1, k}=a_{n, k-1}+a_{n-1, k}
$$
with $n+k-1=n-1+k=Y$ and so we can use induction to obtain
$$
a_{n, k}=frac{(k+n-3)!}{(k-2)!(n-1)!}+frac{(k+n-3)!}{(k-1)!(n-2)!}=frac{(k+n-3)!}{(k-2)!(n-2)!(n-1)}+frac{(k+n-3)!}{(k-2)!(n-2)!(k-1)}=frac{(k+n-3)!(k-1)+(k+n-3)!(n-1)}{(k-2)!(n-2)!(k-1)(n-1)}=frac{(k+n-3)!(k+n-2)}{(k-1)!(n-1)!}=frac{(k+n-2)!}{(k-1)!(n-1)!}
$$
Obtaining so the initial statement.
answered 2 days ago
P De Donato
2236
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We can work by induction on the quantity $Y=k+n$. If $Y=2$ then $k=1$ and $n=1$ so
$$
a_{1, 1}=1=frac{(1+1-2)!}{(1-1)!(1-1)!}
$$
Now suppose that for a certain $Y$ the equality holds for every $k, n$ such that $Y=k+n$, so now set $n, k$ such that $n+k=Y+1$. If $n=1$ then
$$
a_{1, k}=frac{(k-1)!}{(k-1)!}=1
$$
Otherwise if $n>1$ we have
$$
a_{n, k}=a_{n-1, 1}+a_{n-1, 2}+dotsb+a_{n-1, k-1}+a_{n-1, k}=a_{n, k-1}+a_{n-1, k}
$$
with $n+k-1=n-1+k=Y$ and so we can use induction to obtain
$$
a_{n, k}=frac{(k+n-3)!}{(k-2)!(n-1)!}+frac{(k+n-3)!}{(k-1)!(n-2)!}=frac{(k+n-3)!}{(k-2)!(n-2)!(n-1)}+frac{(k+n-3)!}{(k-2)!(n-2)!(k-1)}=frac{(k+n-3)!(k-1)+(k+n-3)!(n-1)}{(k-2)!(n-2)!(k-1)(n-1)}=frac{(k+n-3)!(k+n-2)}{(k-1)!(n-1)!}=frac{(k+n-2)!}{(k-1)!(n-1)!}
$$
Obtaining so the initial statement.
answered 2 days ago
P De Donato
2236
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
P De Donato
2236
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
P De Donato
2236
answered 2 days ago
P De Donato
2236
2236
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
P De Donato is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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