Proof of $sum_i=1^nicdot(n-i) = binomn+13$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to begin combinatorial proof of $sum_k=1^n k binom nk^2 = n binom2n-1n-1$Proof strategy to show $large sum_0 le x le a normalsize binomax binombn+x = binoma+ba+n $Combinatorial proof of identity $sum_k=0^min(a,b)binomx+y+kkbinomxb-kbinomya-k = binomx+abbinomy+ba$Combinatorial Proof for the Identity $sum_k=0^n binomnkleft(-2right)^k = (-1)^n$Combinatorial proof of $binomkibinomnk=binomnibinomn-ik-i$Combinatorial Proof of $n(n+1)2^n-2 = sum_i=1^ni^2binomni$Combinatorial Proof of $binomn+mk = sum_i=0^k binomni binommk-i$Method for approaching combinatorial proofsCombinatorial proof of $binom3n3 =3binomn3 +6nbinomn2 +n^3$?Combinatorial proof of $sum_k=1^n k^2 =binomn+13 + binomn+23$
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Proof of $sum_i=1^nicdot(n-i) = binomn+13$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to begin combinatorial proof of $sum_k=1^n k binom nk^2 = n binom2n-1n-1$Proof strategy to show $large sum_0 le x le a normalsize binomax binombn+x = binoma+ba+n $Combinatorial proof of identity $sum_k=0^min(a,b)binomx+y+kkbinomxb-kbinomya-k = binomx+abbinomy+ba$Combinatorial Proof for the Identity $sum_k=0^n binomnkleft(-2right)^k = (-1)^n$Combinatorial proof of $binomkibinomnk=binomnibinomn-ik-i$Combinatorial Proof of $n(n+1)2^n-2 = sum_i=1^ni^2binomni$Combinatorial Proof of $binomn+mk = sum_i=0^k binomni binommk-i$Method for approaching combinatorial proofsCombinatorial proof of $binom3n3 =3binomn3 +6nbinomn2 +n^3$?Combinatorial proof of $sum_k=1^n k^2 =binomn+13 + binomn+23$
$begingroup$
I would like to prove combinatorially that $sum_i=1^nicdot(n-i) = binomn+13$.
Algebraically, this identity is easily proved in the following way:
$LHS = (1+2+cdotcdotcdot+n-1+n)cdot n - (1^2+2^2+cdotcdotcdot+n^2)\=n^2(n+1)/2 - n(n+1)(2n+1)/6 = (n+1)(n-1)cdot n/6=RHS$
However, is there any combinatorial proof for this equality?
combinatorics discrete-mathematics
edited Apr 8 at 22:18
The Count
2,32361431
asked Sep 14 '17 at 5:49
BeverlieBeverlie
1,158323
$endgroup$
add a comment |
$begingroup$
I would like to prove combinatorially that $sum_i=1^nicdot(n-i) = binomn+13$.
Algebraically, this identity is easily proved in the following way:
$LHS = (1+2+cdotcdotcdot+n-1+n)cdot n - (1^2+2^2+cdotcdotcdot+n^2)\=n^2(n+1)/2 - n(n+1)(2n+1)/6 = (n+1)(n-1)cdot n/6=RHS$
However, is there any combinatorial proof for this equality?
combinatorics discrete-mathematics
edited Apr 8 at 22:18
The Count
2,32361431
asked Sep 14 '17 at 5:49
BeverlieBeverlie
1,158323
$endgroup$
$begingroup$
What is "Combinatoric way"?
$endgroup$
– Krish
Sep 14 '17 at 5:51
$begingroup$
@Krish OP is presumably asking for a combinatorial proof of the identity.
$endgroup$
– JMoravitz
Sep 14 '17 at 5:51
$begingroup$
@JMoravitz thanks edited OP
$endgroup$
– Beverlie
Sep 14 '17 at 5:52
add a comment |
$begingroup$
I would like to prove combinatorially that $sum_i=1^nicdot(n-i) = binomn+13$.
Algebraically, this identity is easily proved in the following way:
$LHS = (1+2+cdotcdotcdot+n-1+n)cdot n - (1^2+2^2+cdotcdotcdot+n^2)\=n^2(n+1)/2 - n(n+1)(2n+1)/6 = (n+1)(n-1)cdot n/6=RHS$
However, is there any combinatorial proof for this equality?
combinatorics discrete-mathematics
edited Apr 8 at 22:18
The Count
2,32361431
asked Sep 14 '17 at 5:49
BeverlieBeverlie
1,158323
$endgroup$
I would like to prove combinatorially that $sum_i=1^nicdot(n-i) = binomn+13$.
Algebraically, this identity is easily proved in the following way:
$LHS = (1+2+cdotcdotcdot+n-1+n)cdot n - (1^2+2^2+cdotcdotcdot+n^2)\=n^2(n+1)/2 - n(n+1)(2n+1)/6 = (n+1)(n-1)cdot n/6=RHS$
However, is there any combinatorial proof for this equality?
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited Apr 8 at 22:18
The Count
2,32361431
asked Sep 14 '17 at 5:49
BeverlieBeverlie
1,158323
edited Apr 8 at 22:18
The Count
2,32361431
asked Sep 14 '17 at 5:49
BeverlieBeverlie
1,158323
edited Apr 8 at 22:18
The Count
2,32361431
edited Apr 8 at 22:18
The Count
2,32361431
edited Apr 8 at 22:18
The Count
2,32361431
2,32361431
asked Sep 14 '17 at 5:49
BeverlieBeverlie
1,158323
asked Sep 14 '17 at 5:49
BeverlieBeverlie
1,158323
asked Sep 14 '17 at 5:49
BeverlieBeverlie
1,158323
1,158323
$begingroup$
What is "Combinatoric way"?
$endgroup$
– Krish
Sep 14 '17 at 5:51
$begingroup$
@Krish OP is presumably asking for a combinatorial proof of the identity.
$endgroup$
– JMoravitz
Sep 14 '17 at 5:51
$begingroup$
@JMoravitz thanks edited OP
$endgroup$
– Beverlie
Sep 14 '17 at 5:52
add a comment |
$begingroup$
What is "Combinatoric way"?
$endgroup$
– Krish
Sep 14 '17 at 5:51
$begingroup$
@Krish OP is presumably asking for a combinatorial proof of the identity.
$endgroup$
– JMoravitz
Sep 14 '17 at 5:51
$begingroup$
@JMoravitz thanks edited OP
$endgroup$
– Beverlie
Sep 14 '17 at 5:52
$begingroup$
What is "Combinatoric way"?
$endgroup$
– Krish
Sep 14 '17 at 5:51
$begingroup$
What is "Combinatoric way"?
$endgroup$
– Krish
Sep 14 '17 at 5:51
$begingroup$
@Krish OP is presumably asking for a combinatorial proof of the identity.
$endgroup$
– JMoravitz
Sep 14 '17 at 5:51
$begingroup$
@Krish OP is presumably asking for a combinatorial proof of the identity.
$endgroup$
– JMoravitz
Sep 14 '17 at 5:51
$begingroup$
@JMoravitz thanks edited OP
$endgroup$
– Beverlie
Sep 14 '17 at 5:52
$begingroup$
@JMoravitz thanks edited OP
$endgroup$
– Beverlie
Sep 14 '17 at 5:52
add a comment |
1 Answer
1
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$begingroup$
Consider the three-element subsets of $0,1,ldots,n$. There are
$binomn+13$ of them. How many have "middle element" $i$?
If the set has middle element $i$, there are $i$ choices for the
smallest element and $n-i$ choices for the largest element, so there are
$i(n-i)$ three-element subsets of $0,1,ldots,n$ with middle element $i$.
answered Sep 14 '17 at 5:53
Lord Shark the UnknownLord Shark the Unknown
108k1163136
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
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active
oldest
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active
oldest
votes
$begingroup$
Consider the three-element subsets of $0,1,ldots,n$. There are
$binomn+13$ of them. How many have "middle element" $i$?
If the set has middle element $i$, there are $i$ choices for the
smallest element and $n-i$ choices for the largest element, so there are
$i(n-i)$ three-element subsets of $0,1,ldots,n$ with middle element $i$.
answered Sep 14 '17 at 5:53
Lord Shark the UnknownLord Shark the Unknown
108k1163136
$endgroup$
add a comment |
$begingroup$
Consider the three-element subsets of $0,1,ldots,n$. There are
$binomn+13$ of them. How many have "middle element" $i$?
If the set has middle element $i$, there are $i$ choices for the
smallest element and $n-i$ choices for the largest element, so there are
$i(n-i)$ three-element subsets of $0,1,ldots,n$ with middle element $i$.
answered Sep 14 '17 at 5:53
Lord Shark the UnknownLord Shark the Unknown
108k1163136
$endgroup$
add a comment |
$begingroup$
Consider the three-element subsets of $0,1,ldots,n$. There are
$binomn+13$ of them. How many have "middle element" $i$?
If the set has middle element $i$, there are $i$ choices for the
smallest element and $n-i$ choices for the largest element, so there are
$i(n-i)$ three-element subsets of $0,1,ldots,n$ with middle element $i$.
answered Sep 14 '17 at 5:53
Lord Shark the UnknownLord Shark the Unknown
108k1163136
$endgroup$
Consider the three-element subsets of $0,1,ldots,n$. There are
$binomn+13$ of them. How many have "middle element" $i$?
If the set has middle element $i$, there are $i$ choices for the
smallest element and $n-i$ choices for the largest element, so there are
$i(n-i)$ three-element subsets of $0,1,ldots,n$ with middle element $i$.
answered Sep 14 '17 at 5:53
Lord Shark the UnknownLord Shark the Unknown
108k1163136
answered Sep 14 '17 at 5:53
Lord Shark the UnknownLord Shark the Unknown
108k1163136
answered Sep 14 '17 at 5:53
Lord Shark the UnknownLord Shark the Unknown
108k1163136
answered Sep 14 '17 at 5:53
Lord Shark the UnknownLord Shark the Unknown
108k1163136
108k1163136
add a comment |
add a comment |
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$begingroup$
What is "Combinatoric way"?
$endgroup$
– Krish
Sep 14 '17 at 5:51
$begingroup$
@Krish OP is presumably asking for a combinatorial proof of the identity.
$endgroup$
– JMoravitz
Sep 14 '17 at 5:51
$begingroup$
@JMoravitz thanks edited OP
$endgroup$
– Beverlie
Sep 14 '17 at 5:52