Proving $sum_m=0^nbinomnm^2 binommn-k=binomnkbinomn+kk$ The 2019 Stack Overflow Developer Survey Results Are InHow to prove that $sum_k=0^n binom nk k^2=2^n-2(n^2+n)$Evaluate $sum_r=0^n binomnrsin rx cos (n-r)x$Prove that $sum_r=1^n frac 1rbinomnr = sum_r=1^n frac 1r(2^r - 1)$Prove $sum_0^n(-1)^ibinomni = 0$Sum of combinatorial series $binom6n3r$ where r varies from 0 to nFind the sum of series: $sum_k=0^n (-1)^k binomnk binom2n-kn$Proving this infinite sum of a product of three binomials: $sumlimits_sbinomn+sk+lbinom ksbinom ls=binom nkbinom nl$Prove $sum_r=0^n binomnr binomn+rr (-2)^r =(-1)^nsum_r=0^n binomnr^2 2^r$Prove that $sum_k = 0^49(-1)^kbinom992k = -2^49$even number - binomial
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Proving $sum_m=0^nbinomnm^2 binommn-k=binomnkbinomn+kk$
The 2019 Stack Overflow Developer Survey Results Are InHow to prove that $sum_k=0^n binom nk k^2=2^n-2(n^2+n)$Evaluate $sum_r=0^n binomnrsin rx cos (n-r)x$Prove that $sum_r=1^n frac 1rbinomnr = sum_r=1^n frac 1r(2^r - 1)$Prove $sum_0^n(-1)^ibinomni = 0$Sum of combinatorial series $binom6n3r$ where r varies from 0 to nFind the sum of series: $sum_k=0^n (-1)^k binomnk binom2n-kn$Proving this infinite sum of a product of three binomials: $sumlimits_sbinomn+sk+lbinom ksbinom ls=binom nkbinom nl$Prove $sum_r=0^n binomnr binomn+rr (-2)^r =(-1)^nsum_r=0^n binomnr^2 2^r$Prove that $sum_k = 0^49(-1)^kbinom992k = -2^49$even number - binomial
$begingroup$
How can I prove this?
$$sum_m=0^nbinomnm^2 binommn-k=binomnkbinomn+kk$$
$$ 0le k le n $$
I developed the expressions, but they are not the same. I do not know if it will be my mistake or the advisor
I tried to solve for the binomial, but I could not, any idea to be able to proceed
binomial-coefficients binomial-theorem
edited Mar 30 at 11:59
Blue
49.5k870158
asked Mar 30 at 9:24
MonicaMonica
215
$endgroup$
add a comment |
$begingroup$
How can I prove this?
$$sum_m=0^nbinomnm^2 binommn-k=binomnkbinomn+kk$$
$$ 0le k le n $$
I developed the expressions, but they are not the same. I do not know if it will be my mistake or the advisor
I tried to solve for the binomial, but I could not, any idea to be able to proceed
binomial-coefficients binomial-theorem
edited Mar 30 at 11:59
Blue
49.5k870158
asked Mar 30 at 9:24
MonicaMonica
215
$endgroup$
$begingroup$
I think that they are the same. Try with small numbers to check
$endgroup$
– Claude Leibovici
Mar 30 at 9:31
$begingroup$
How can I prove it?
$endgroup$
– Monica
Mar 30 at 9:34
add a comment |
$begingroup$
How can I prove this?
$$sum_m=0^nbinomnm^2 binommn-k=binomnkbinomn+kk$$
$$ 0le k le n $$
I developed the expressions, but they are not the same. I do not know if it will be my mistake or the advisor
I tried to solve for the binomial, but I could not, any idea to be able to proceed
binomial-coefficients binomial-theorem
edited Mar 30 at 11:59
Blue
49.5k870158
asked Mar 30 at 9:24
MonicaMonica
215
$endgroup$
How can I prove this?
$$sum_m=0^nbinomnm^2 binommn-k=binomnkbinomn+kk$$
$$ 0le k le n $$
I developed the expressions, but they are not the same. I do not know if it will be my mistake or the advisor
I tried to solve for the binomial, but I could not, any idea to be able to proceed
binomial-coefficients binomial-theorem
binomial-coefficients binomial-theorem
edited Mar 30 at 11:59
Blue
49.5k870158
asked Mar 30 at 9:24
MonicaMonica
215
edited Mar 30 at 11:59
Blue
49.5k870158
asked Mar 30 at 9:24
MonicaMonica
215
edited Mar 30 at 11:59
Blue
49.5k870158
edited Mar 30 at 11:59
Blue
49.5k870158
edited Mar 30 at 11:59
Blue
49.5k870158
49.5k870158
asked Mar 30 at 9:24
MonicaMonica
215
asked Mar 30 at 9:24
MonicaMonica
215
asked Mar 30 at 9:24
MonicaMonica
215
215
$begingroup$
I think that they are the same. Try with small numbers to check
$endgroup$
– Claude Leibovici
Mar 30 at 9:31
$begingroup$
How can I prove it?
$endgroup$
– Monica
Mar 30 at 9:34
add a comment |
$begingroup$
I think that they are the same. Try with small numbers to check
$endgroup$
– Claude Leibovici
Mar 30 at 9:31
$begingroup$
How can I prove it?
$endgroup$
– Monica
Mar 30 at 9:34
$begingroup$
I think that they are the same. Try with small numbers to check
$endgroup$
– Claude Leibovici
Mar 30 at 9:31
$begingroup$
I think that they are the same. Try with small numbers to check
$endgroup$
– Claude Leibovici
Mar 30 at 9:31
$begingroup$
How can I prove it?
$endgroup$
– Monica
Mar 30 at 9:34
$begingroup$
How can I prove it?
$endgroup$
– Monica
Mar 30 at 9:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let us prove the general form of the identity given by @Richard.
$$sum_r=0^infty binom ar binom br binom rc = binom ac binoma+b-ca.$$
Replacing $binom ar binom rc$ with:
$$
binom ar binom rc=fraca!r!(a-r)!fracr!c!(r-c)!
=fraca!c!(a-c)!frac(a-c)!(a-r)!(r-c)!=binom acbinom a-ca-r,
$$
one obtains:
$$sum_r=0^infty binom ar binom br binom rc =
binom acsum_r=0^infty binom br binom a-ca-r
=binom ac binoma+b-ca,
$$
where in the last equality the Vandermonde's identity was used.
Inserting $a=b=n$ and $c=n-k$ yields the identity of OP, as already was pointed out by @Richard.
answered Mar 30 at 11:42
useruser
6,41811031
$endgroup$
$begingroup$
Didn't know this identity was named after Vandermonde, and much simpler proof than mine. 1:0 for you, @user. +1
$endgroup$
– Richard
Mar 30 at 11:58
add a comment |
$begingroup$
$$sum_m=0^nbinomnm^2binommn-k=binomnksum_m=0^nbinomnmbinomkn-m=binomnkbinomn+kn$$where the second equality rests on the vandermonde identity.
answered Mar 30 at 11:20
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
You can use the general identity $$sum_kge0binom akbinom bkbinom kc=binom acbinoma+b-ca$$ Inserting $n$ for $a$ and $b$ and $n-k$ for $c$ yields what you wanted to prove.
The upper bound of summation can be reduced from $infty$ to $n$ because $binom nk$ is 0 for $kgt n$.
EDIT: This can be proven using Vandermonde's identity. @user gives a very simple proof in his answer.
answered Mar 30 at 10:07
RichardRichard
3751211
$endgroup$
1
$begingroup$
It would be nice to give a proof of the general identity or at least a reference to it.
$endgroup$
– user
Mar 30 at 10:43
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us prove the general form of the identity given by @Richard.
$$sum_r=0^infty binom ar binom br binom rc = binom ac binoma+b-ca.$$
Replacing $binom ar binom rc$ with:
$$
binom ar binom rc=fraca!r!(a-r)!fracr!c!(r-c)!
=fraca!c!(a-c)!frac(a-c)!(a-r)!(r-c)!=binom acbinom a-ca-r,
$$
one obtains:
$$sum_r=0^infty binom ar binom br binom rc =
binom acsum_r=0^infty binom br binom a-ca-r
=binom ac binoma+b-ca,
$$
where in the last equality the Vandermonde's identity was used.
Inserting $a=b=n$ and $c=n-k$ yields the identity of OP, as already was pointed out by @Richard.
answered Mar 30 at 11:42
useruser
6,41811031
$endgroup$
$begingroup$
Didn't know this identity was named after Vandermonde, and much simpler proof than mine. 1:0 for you, @user. +1
$endgroup$
– Richard
Mar 30 at 11:58
add a comment |
$begingroup$
Let us prove the general form of the identity given by @Richard.
$$sum_r=0^infty binom ar binom br binom rc = binom ac binoma+b-ca.$$
Replacing $binom ar binom rc$ with:
$$
binom ar binom rc=fraca!r!(a-r)!fracr!c!(r-c)!
=fraca!c!(a-c)!frac(a-c)!(a-r)!(r-c)!=binom acbinom a-ca-r,
$$
one obtains:
$$sum_r=0^infty binom ar binom br binom rc =
binom acsum_r=0^infty binom br binom a-ca-r
=binom ac binoma+b-ca,
$$
where in the last equality the Vandermonde's identity was used.
Inserting $a=b=n$ and $c=n-k$ yields the identity of OP, as already was pointed out by @Richard.
answered Mar 30 at 11:42
useruser
6,41811031
$endgroup$
$begingroup$
Didn't know this identity was named after Vandermonde, and much simpler proof than mine. 1:0 for you, @user. +1
$endgroup$
– Richard
Mar 30 at 11:58
add a comment |
$begingroup$
Let us prove the general form of the identity given by @Richard.
$$sum_r=0^infty binom ar binom br binom rc = binom ac binoma+b-ca.$$
Replacing $binom ar binom rc$ with:
$$
binom ar binom rc=fraca!r!(a-r)!fracr!c!(r-c)!
=fraca!c!(a-c)!frac(a-c)!(a-r)!(r-c)!=binom acbinom a-ca-r,
$$
one obtains:
$$sum_r=0^infty binom ar binom br binom rc =
binom acsum_r=0^infty binom br binom a-ca-r
=binom ac binoma+b-ca,
$$
where in the last equality the Vandermonde's identity was used.
Inserting $a=b=n$ and $c=n-k$ yields the identity of OP, as already was pointed out by @Richard.
answered Mar 30 at 11:42
useruser
6,41811031
$endgroup$
Let us prove the general form of the identity given by @Richard.
$$sum_r=0^infty binom ar binom br binom rc = binom ac binoma+b-ca.$$
Replacing $binom ar binom rc$ with:
$$
binom ar binom rc=fraca!r!(a-r)!fracr!c!(r-c)!
=fraca!c!(a-c)!frac(a-c)!(a-r)!(r-c)!=binom acbinom a-ca-r,
$$
one obtains:
$$sum_r=0^infty binom ar binom br binom rc =
binom acsum_r=0^infty binom br binom a-ca-r
=binom ac binoma+b-ca,
$$
where in the last equality the Vandermonde's identity was used.
Inserting $a=b=n$ and $c=n-k$ yields the identity of OP, as already was pointed out by @Richard.
answered Mar 30 at 11:42
useruser
6,41811031
edited Mar 30 at 15:41
answered Mar 30 at 11:42
useruser
6,41811031
answered Mar 30 at 11:42
useruser
6,41811031
answered Mar 30 at 11:42
useruser
6,41811031
6,41811031
$begingroup$
Didn't know this identity was named after Vandermonde, and much simpler proof than mine. 1:0 for you, @user. +1
$endgroup$
– Richard
Mar 30 at 11:58
add a comment |
$begingroup$
Didn't know this identity was named after Vandermonde, and much simpler proof than mine. 1:0 for you, @user. +1
$endgroup$
– Richard
Mar 30 at 11:58
$begingroup$
Didn't know this identity was named after Vandermonde, and much simpler proof than mine. 1:0 for you, @user. +1
$endgroup$
– Richard
Mar 30 at 11:58
$begingroup$
Didn't know this identity was named after Vandermonde, and much simpler proof than mine. 1:0 for you, @user. +1
$endgroup$
– Richard
Mar 30 at 11:58
add a comment |
$begingroup$
$$sum_m=0^nbinomnm^2binommn-k=binomnksum_m=0^nbinomnmbinomkn-m=binomnkbinomn+kn$$where the second equality rests on the vandermonde identity.
answered Mar 30 at 11:20
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
$$sum_m=0^nbinomnm^2binommn-k=binomnksum_m=0^nbinomnmbinomkn-m=binomnkbinomn+kn$$where the second equality rests on the vandermonde identity.
answered Mar 30 at 11:20
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
$$sum_m=0^nbinomnm^2binommn-k=binomnksum_m=0^nbinomnmbinomkn-m=binomnkbinomn+kn$$where the second equality rests on the vandermonde identity.
answered Mar 30 at 11:20
drhabdrhab
104k545136
$endgroup$
$$sum_m=0^nbinomnm^2binommn-k=binomnksum_m=0^nbinomnmbinomkn-m=binomnkbinomn+kn$$where the second equality rests on the vandermonde identity.
answered Mar 30 at 11:20
drhabdrhab
104k545136
answered Mar 30 at 11:20
drhabdrhab
104k545136
answered Mar 30 at 11:20
drhabdrhab
104k545136
answered Mar 30 at 11:20
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
$begingroup$
You can use the general identity $$sum_kge0binom akbinom bkbinom kc=binom acbinoma+b-ca$$ Inserting $n$ for $a$ and $b$ and $n-k$ for $c$ yields what you wanted to prove.
The upper bound of summation can be reduced from $infty$ to $n$ because $binom nk$ is 0 for $kgt n$.
EDIT: This can be proven using Vandermonde's identity. @user gives a very simple proof in his answer.
answered Mar 30 at 10:07
RichardRichard
3751211
$endgroup$
1
$begingroup$
It would be nice to give a proof of the general identity or at least a reference to it.
$endgroup$
– user
Mar 30 at 10:43
add a comment |
$begingroup$
You can use the general identity $$sum_kge0binom akbinom bkbinom kc=binom acbinoma+b-ca$$ Inserting $n$ for $a$ and $b$ and $n-k$ for $c$ yields what you wanted to prove.
The upper bound of summation can be reduced from $infty$ to $n$ because $binom nk$ is 0 for $kgt n$.
EDIT: This can be proven using Vandermonde's identity. @user gives a very simple proof in his answer.
answered Mar 30 at 10:07
RichardRichard
3751211
$endgroup$
1
$begingroup$
It would be nice to give a proof of the general identity or at least a reference to it.
$endgroup$
– user
Mar 30 at 10:43
add a comment |
$begingroup$
You can use the general identity $$sum_kge0binom akbinom bkbinom kc=binom acbinoma+b-ca$$ Inserting $n$ for $a$ and $b$ and $n-k$ for $c$ yields what you wanted to prove.
The upper bound of summation can be reduced from $infty$ to $n$ because $binom nk$ is 0 for $kgt n$.
EDIT: This can be proven using Vandermonde's identity. @user gives a very simple proof in his answer.
answered Mar 30 at 10:07
RichardRichard
3751211
$endgroup$
You can use the general identity $$sum_kge0binom akbinom bkbinom kc=binom acbinoma+b-ca$$ Inserting $n$ for $a$ and $b$ and $n-k$ for $c$ yields what you wanted to prove.
The upper bound of summation can be reduced from $infty$ to $n$ because $binom nk$ is 0 for $kgt n$.
EDIT: This can be proven using Vandermonde's identity. @user gives a very simple proof in his answer.
answered Mar 30 at 10:07
RichardRichard
3751211
edited Mar 30 at 11:49
answered Mar 30 at 10:07
RichardRichard
3751211
answered Mar 30 at 10:07
RichardRichard
3751211
answered Mar 30 at 10:07
RichardRichard
3751211
3751211
1
$begingroup$
It would be nice to give a proof of the general identity or at least a reference to it.
$endgroup$
– user
Mar 30 at 10:43
add a comment |
1
$begingroup$
It would be nice to give a proof of the general identity or at least a reference to it.
$endgroup$
– user
Mar 30 at 10:43
1
1
$begingroup$
It would be nice to give a proof of the general identity or at least a reference to it.
$endgroup$
– user
Mar 30 at 10:43
$begingroup$
It would be nice to give a proof of the general identity or at least a reference to it.
$endgroup$
– user
Mar 30 at 10:43
add a comment |
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$begingroup$
I think that they are the same. Try with small numbers to check
$endgroup$
– Claude Leibovici
Mar 30 at 9:31
$begingroup$
How can I prove it?
$endgroup$
– Monica
Mar 30 at 9:34