How to prove $sumlimits_{i=0}^{k}(-1)^ibinom{n}{k-i}binom{n+i-1}{i}=0$
$begingroup$
I saw a combinatorial identity when i study linear-algebra, But the author didn't explain how to get it.
$displaystyle sum_{i=0}^{k}(-1)^ibinom{n}{k-i}binom{n+i-1}{i}=0$
Who can help me to prove it? Thanks!
combinatorics binomial-coefficients
edited Sep 5 '16 at 18:55
Did
249k23229468
asked Sep 2 '15 at 11:15
smallsmallicesmallsmallice
438213
$endgroup$
add a comment |
$begingroup$
I saw a combinatorial identity when i study linear-algebra, But the author didn't explain how to get it.
$displaystyle sum_{i=0}^{k}(-1)^ibinom{n}{k-i}binom{n+i-1}{i}=0$
Who can help me to prove it? Thanks!
combinatorics binomial-coefficients
edited Sep 5 '16 at 18:55
Did
249k23229468
asked Sep 2 '15 at 11:15
smallsmallicesmallsmallice
438213
$endgroup$
$begingroup$
I think it would be convenient if you please elaborate on the assignment of the variables. Is i is of complex number?
$endgroup$
– Aneek
Sep 4 '15 at 15:51
$begingroup$
oh ,no.i is also natural number as n,k.
$endgroup$
– smallsmallice
Sep 5 '15 at 6:41
$begingroup$
oh I see. Thanks!
$endgroup$
– Aneek
Sep 5 '15 at 10:16
add a comment |
$begingroup$
I saw a combinatorial identity when i study linear-algebra, But the author didn't explain how to get it.
$displaystyle sum_{i=0}^{k}(-1)^ibinom{n}{k-i}binom{n+i-1}{i}=0$
Who can help me to prove it? Thanks!
combinatorics binomial-coefficients
edited Sep 5 '16 at 18:55
Did
249k23229468
asked Sep 2 '15 at 11:15
smallsmallicesmallsmallice
438213
$endgroup$
I saw a combinatorial identity when i study linear-algebra, But the author didn't explain how to get it.
$displaystyle sum_{i=0}^{k}(-1)^ibinom{n}{k-i}binom{n+i-1}{i}=0$
Who can help me to prove it? Thanks!
combinatorics binomial-coefficients
combinatorics binomial-coefficients
edited Sep 5 '16 at 18:55
Did
249k23229468
asked Sep 2 '15 at 11:15
smallsmallicesmallsmallice
438213
edited Sep 5 '16 at 18:55
Did
249k23229468
asked Sep 2 '15 at 11:15
smallsmallicesmallsmallice
438213
edited Sep 5 '16 at 18:55
Did
249k23229468
edited Sep 5 '16 at 18:55
Did
249k23229468
edited Sep 5 '16 at 18:55
Did
249k23229468
249k23229468
asked Sep 2 '15 at 11:15
smallsmallicesmallsmallice
438213
asked Sep 2 '15 at 11:15
smallsmallicesmallsmallice
438213
asked Sep 2 '15 at 11:15
smallsmallicesmallsmallice
438213
438213
$begingroup$
I think it would be convenient if you please elaborate on the assignment of the variables. Is i is of complex number?
$endgroup$
– Aneek
Sep 4 '15 at 15:51
$begingroup$
oh ,no.i is also natural number as n,k.
$endgroup$
– smallsmallice
Sep 5 '15 at 6:41
$begingroup$
oh I see. Thanks!
$endgroup$
– Aneek
Sep 5 '15 at 10:16
add a comment |
$begingroup$
I think it would be convenient if you please elaborate on the assignment of the variables. Is i is of complex number?
$endgroup$
– Aneek
Sep 4 '15 at 15:51
$begingroup$
oh ,no.i is also natural number as n,k.
$endgroup$
– smallsmallice
Sep 5 '15 at 6:41
$begingroup$
oh I see. Thanks!
$endgroup$
– Aneek
Sep 5 '15 at 10:16
$begingroup$
I think it would be convenient if you please elaborate on the assignment of the variables. Is i is of complex number?
$endgroup$
– Aneek
Sep 4 '15 at 15:51
$begingroup$
I think it would be convenient if you please elaborate on the assignment of the variables. Is i is of complex number?
$endgroup$
– Aneek
Sep 4 '15 at 15:51
$begingroup$
oh ,no.i is also natural number as n,k.
$endgroup$
– smallsmallice
Sep 5 '15 at 6:41
$begingroup$
oh ,no.i is also natural number as n,k.
$endgroup$
– smallsmallice
Sep 5 '15 at 6:41
$begingroup$
oh I see. Thanks!
$endgroup$
– Aneek
Sep 5 '15 at 10:16
$begingroup$
oh I see. Thanks!
$endgroup$
– Aneek
Sep 5 '15 at 10:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You probably want to require $k > 0$, as otherwise your claim is $1 = 0$.
Renaming $k$, $n$ and $i$ as $n$, $x$ and $k$, you get a particular case (the case $x=y$) of the following identity:
$sumlimits_{k=0}^n left(-1right)^k dbinom{x}{n-k} dbinom{k+y-1}{k} = dbinom{x-y}{n}$.
This is, e.g., Proposition 3.32 (d) in my Notes on the combinatorial
fundamentals of algebra. (In case of changing numbering, search for " some sample applications of Theorem", or check out the frozen version of 10 January 2019, in which the numbering surely has not shifted.) The main trick to the proof is realizing that $left(-1right)^k dbinom{k+y-1}{k}$ can be rewritten as $dbinom{-y}{k}$, after which you are left with a particular case of the Vandermonde convolution identity.
answered Sep 2 '15 at 11:27
darij grinbergdarij grinberg
11.6k33168
$endgroup$
$begingroup$
Pretty good book！but seems a little difficult to me ... I' ll try my best to read it ! Thank you very much !
$endgroup$
– smallsmallice
Sep 2 '15 at 12:14
$begingroup$
ok~thanks a lot!
$endgroup$
– smallsmallice
Sep 2 '15 at 13:58
$begingroup$
This isn't a book, though it might become one... The proof of Proposition 2.25 is mostly self-contained, relying only on Theorem 2.24 and on some basic identities such as upper negation (Proposition 2.16). If you have trouble with the notations (such as binomial coefficients $dbinom{m}{n}$ with negative $m$), look them up at the beginning of Chapter 2.
$endgroup$
– darij grinberg
Sep 5 '16 at 17:45
add a comment |
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1 Answer
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$begingroup$
You probably want to require $k > 0$, as otherwise your claim is $1 = 0$.
Renaming $k$, $n$ and $i$ as $n$, $x$ and $k$, you get a particular case (the case $x=y$) of the following identity:
$sumlimits_{k=0}^n left(-1right)^k dbinom{x}{n-k} dbinom{k+y-1}{k} = dbinom{x-y}{n}$.
This is, e.g., Proposition 3.32 (d) in my Notes on the combinatorial
fundamentals of algebra. (In case of changing numbering, search for " some sample applications of Theorem", or check out the frozen version of 10 January 2019, in which the numbering surely has not shifted.) The main trick to the proof is realizing that $left(-1right)^k dbinom{k+y-1}{k}$ can be rewritten as $dbinom{-y}{k}$, after which you are left with a particular case of the Vandermonde convolution identity.
answered Sep 2 '15 at 11:27
darij grinbergdarij grinberg
11.6k33168
$endgroup$
$begingroup$
Pretty good book！but seems a little difficult to me ... I' ll try my best to read it ! Thank you very much !
$endgroup$
– smallsmallice
Sep 2 '15 at 12:14
$begingroup$
ok~thanks a lot!
$endgroup$
– smallsmallice
Sep 2 '15 at 13:58
$begingroup$
This isn't a book, though it might become one... The proof of Proposition 2.25 is mostly self-contained, relying only on Theorem 2.24 and on some basic identities such as upper negation (Proposition 2.16). If you have trouble with the notations (such as binomial coefficients $dbinom{m}{n}$ with negative $m$), look them up at the beginning of Chapter 2.
$endgroup$
– darij grinberg
Sep 5 '16 at 17:45
add a comment |
$begingroup$
You probably want to require $k > 0$, as otherwise your claim is $1 = 0$.
Renaming $k$, $n$ and $i$ as $n$, $x$ and $k$, you get a particular case (the case $x=y$) of the following identity:
$sumlimits_{k=0}^n left(-1right)^k dbinom{x}{n-k} dbinom{k+y-1}{k} = dbinom{x-y}{n}$.
This is, e.g., Proposition 3.32 (d) in my Notes on the combinatorial
fundamentals of algebra. (In case of changing numbering, search for " some sample applications of Theorem", or check out the frozen version of 10 January 2019, in which the numbering surely has not shifted.) The main trick to the proof is realizing that $left(-1right)^k dbinom{k+y-1}{k}$ can be rewritten as $dbinom{-y}{k}$, after which you are left with a particular case of the Vandermonde convolution identity.
answered Sep 2 '15 at 11:27
darij grinbergdarij grinberg
11.6k33168
$endgroup$
$begingroup$
Pretty good book！but seems a little difficult to me ... I' ll try my best to read it ! Thank you very much !
$endgroup$
– smallsmallice
Sep 2 '15 at 12:14
$begingroup$
ok~thanks a lot!
$endgroup$
– smallsmallice
Sep 2 '15 at 13:58
$begingroup$
This isn't a book, though it might become one... The proof of Proposition 2.25 is mostly self-contained, relying only on Theorem 2.24 and on some basic identities such as upper negation (Proposition 2.16). If you have trouble with the notations (such as binomial coefficients $dbinom{m}{n}$ with negative $m$), look them up at the beginning of Chapter 2.
$endgroup$
– darij grinberg
Sep 5 '16 at 17:45
add a comment |
$begingroup$
You probably want to require $k > 0$, as otherwise your claim is $1 = 0$.
Renaming $k$, $n$ and $i$ as $n$, $x$ and $k$, you get a particular case (the case $x=y$) of the following identity:
$sumlimits_{k=0}^n left(-1right)^k dbinom{x}{n-k} dbinom{k+y-1}{k} = dbinom{x-y}{n}$.
This is, e.g., Proposition 3.32 (d) in my Notes on the combinatorial
fundamentals of algebra. (In case of changing numbering, search for " some sample applications of Theorem", or check out the frozen version of 10 January 2019, in which the numbering surely has not shifted.) The main trick to the proof is realizing that $left(-1right)^k dbinom{k+y-1}{k}$ can be rewritten as $dbinom{-y}{k}$, after which you are left with a particular case of the Vandermonde convolution identity.
answered Sep 2 '15 at 11:27
darij grinbergdarij grinberg
11.6k33168
$endgroup$
You probably want to require $k > 0$, as otherwise your claim is $1 = 0$.
Renaming $k$, $n$ and $i$ as $n$, $x$ and $k$, you get a particular case (the case $x=y$) of the following identity:
$sumlimits_{k=0}^n left(-1right)^k dbinom{x}{n-k} dbinom{k+y-1}{k} = dbinom{x-y}{n}$.
This is, e.g., Proposition 3.32 (d) in my Notes on the combinatorial
fundamentals of algebra. (In case of changing numbering, search for " some sample applications of Theorem", or check out the frozen version of 10 January 2019, in which the numbering surely has not shifted.) The main trick to the proof is realizing that $left(-1right)^k dbinom{k+y-1}{k}$ can be rewritten as $dbinom{-y}{k}$, after which you are left with a particular case of the Vandermonde convolution identity.
answered Sep 2 '15 at 11:27
darij grinbergdarij grinberg
11.6k33168
edited Jan 15 at 22:24
answered Sep 2 '15 at 11:27
darij grinbergdarij grinberg
11.6k33168
answered Sep 2 '15 at 11:27
darij grinbergdarij grinberg
11.6k33168
answered Sep 2 '15 at 11:27
darij grinbergdarij grinberg
11.6k33168
11.6k33168
$begingroup$
Pretty good book！but seems a little difficult to me ... I' ll try my best to read it ! Thank you very much !
$endgroup$
– smallsmallice
Sep 2 '15 at 12:14
$begingroup$
ok~thanks a lot!
$endgroup$
– smallsmallice
Sep 2 '15 at 13:58
$begingroup$
This isn't a book, though it might become one... The proof of Proposition 2.25 is mostly self-contained, relying only on Theorem 2.24 and on some basic identities such as upper negation (Proposition 2.16). If you have trouble with the notations (such as binomial coefficients $dbinom{m}{n}$ with negative $m$), look them up at the beginning of Chapter 2.
$endgroup$
– darij grinberg
Sep 5 '16 at 17:45
add a comment |
$begingroup$
Pretty good book！but seems a little difficult to me ... I' ll try my best to read it ! Thank you very much !
$endgroup$
– smallsmallice
Sep 2 '15 at 12:14
$begingroup$
ok~thanks a lot!
$endgroup$
– smallsmallice
Sep 2 '15 at 13:58
$begingroup$
This isn't a book, though it might become one... The proof of Proposition 2.25 is mostly self-contained, relying only on Theorem 2.24 and on some basic identities such as upper negation (Proposition 2.16). If you have trouble with the notations (such as binomial coefficients $dbinom{m}{n}$ with negative $m$), look them up at the beginning of Chapter 2.
$endgroup$
– darij grinberg
Sep 5 '16 at 17:45
$begingroup$
Pretty good book！but seems a little difficult to me ... I' ll try my best to read it ! Thank you very much !
$endgroup$
– smallsmallice
Sep 2 '15 at 12:14
$begingroup$
Pretty good book！but seems a little difficult to me ... I' ll try my best to read it ! Thank you very much !
$endgroup$
– smallsmallice
Sep 2 '15 at 12:14
$begingroup$
ok~thanks a lot!
$endgroup$
– smallsmallice
Sep 2 '15 at 13:58
$begingroup$
ok~thanks a lot!
$endgroup$
– smallsmallice
Sep 2 '15 at 13:58
$begingroup$
This isn't a book, though it might become one... The proof of Proposition 2.25 is mostly self-contained, relying only on Theorem 2.24 and on some basic identities such as upper negation (Proposition 2.16). If you have trouble with the notations (such as binomial coefficients $dbinom{m}{n}$ with negative $m$), look them up at the beginning of Chapter 2.
$endgroup$
– darij grinberg
Sep 5 '16 at 17:45
$begingroup$
This isn't a book, though it might become one... The proof of Proposition 2.25 is mostly self-contained, relying only on Theorem 2.24 and on some basic identities such as upper negation (Proposition 2.16). If you have trouble with the notations (such as binomial coefficients $dbinom{m}{n}$ with negative $m$), look them up at the beginning of Chapter 2.
$endgroup$
– darij grinberg
Sep 5 '16 at 17:45
add a comment |
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$begingroup$
I think it would be convenient if you please elaborate on the assignment of the variables. Is i is of complex number?
$endgroup$
– Aneek
Sep 4 '15 at 15:51
$begingroup$
oh ,no.i is also natural number as n,k.
$endgroup$
– smallsmallice
Sep 5 '15 at 6:41
$begingroup$
oh I see. Thanks!
$endgroup$
– Aneek
Sep 5 '15 at 10:16