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Solving summation when $i$ and $j$ are dependent: $sumsum_0 leq i <j leq n1$ [closed]
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How to solve the summation when the indices $i$ and $j$ are dependent?
Please, anybody, explain. I am having no idea how to proceed.
Find the sum $$sum_0 leq i <sum_!!j leq n1$$
algebra-precalculus
edited Jul 20 at 12:34
Nash J.
1,055315
asked Jul 20 at 11:40
Rafael Nadal
1076
closed as unclear what you're asking by Did, Claude Leibovici, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz Jul 25 at 14:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-1
down vote
favorite
How to solve the summation when the indices $i$ and $j$ are dependent?
Please, anybody, explain. I am having no idea how to proceed.
Find the sum $$sum_0 leq i <sum_!!j leq n1$$
algebra-precalculus
edited Jul 20 at 12:34
Nash J.
1,055315
asked Jul 20 at 11:40
Rafael Nadal
1076
closed as unclear what you're asking by Did, Claude Leibovici, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz Jul 25 at 14:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
I am a highschool student who just want explaination please don't close my question
– Rafael Nadal
Jul 20 at 11:42
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How to solve the summation when the indices $i$ and $j$ are dependent?
Please, anybody, explain. I am having no idea how to proceed.
Find the sum $$sum_0 leq i <sum_!!j leq n1$$
algebra-precalculus
edited Jul 20 at 12:34
Nash J.
1,055315
asked Jul 20 at 11:40
Rafael Nadal
1076
How to solve the summation when the indices $i$ and $j$ are dependent?
Please, anybody, explain. I am having no idea how to proceed.
Find the sum $$sum_0 leq i <sum_!!j leq n1$$
algebra-precalculus
edited Jul 20 at 12:34
Nash J.
1,055315
asked Jul 20 at 11:40
Rafael Nadal
1076
edited Jul 20 at 12:34
Nash J.
1,055315
edited Jul 20 at 12:34
Nash J.
1,055315
edited Jul 20 at 12:34
Nash J.
1,055315
1,055315
asked Jul 20 at 11:40
Rafael Nadal
1076
asked Jul 20 at 11:40
Rafael Nadal
1076
asked Jul 20 at 11:40
Rafael Nadal
1076
1076
closed as unclear what you're asking by Did, Claude Leibovici, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz Jul 25 at 14:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, Claude Leibovici, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz Jul 25 at 14:39
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
I am a highschool student who just want explaination please don't close my question
– Rafael Nadal
Jul 20 at 11:42
add a comment |Â
1
I am a highschool student who just want explaination please don't close my question
– Rafael Nadal
Jul 20 at 11:42
1
1
I am a highschool student who just want explaination please don't close my question
– Rafael Nadal
Jul 20 at 11:42
I am a highschool student who just want explaination please don't close my question
– Rafael Nadal
Jul 20 at 11:42
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
The domain of the integer pairs $(i,j)$ such that
$$0le i<jle n$$ is a triangle inside the square of size $n+1$ by $n+1$, diagonal excluded. Summing the term $1$ means that you just count those elements.
Hence you compute the area of the square minus the diagonal, and halve, giving
$$frac(n+1)^2-(n+1)2=fracn(n+1)2.$$
answered Jul 20 at 12:32
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
Note that you can convert the expression into:
$$sum_i=0^n-1bigg(sum_j=i+1^n 1bigg) = sum_i=0^n-1 (n-i) = sum_i=0^n-1 n - sum_i=0^n-1 i = n^2-fracn(n-1)2 = fracn(n+1)2$$
When you see double sums like that, try breaking it up into individual sums. If one index (in this case, $j$) is conditional on another ($i$), the limit(s) of the dependent index can be expressed in terms of the other index. There's no reference/resource as such that comes to my mind, but it should get easier as you solve more and more such problems.
answered Jul 20 at 11:52
Shirish Kulhari
929215
Sir can u provide additional resources regarding this question
– Rafael Nadal
Jul 20 at 11:53
add a comment |Â
up vote
0
down vote
We have that
$$mathopsumsum_0le i<jle n 1=sum_i=0^n-1 left(sum_j=i+1^n 1right)=sum_i=0^n-1 (n-i)=nsum_i=0^n-11-sum_i=0^n-1 i=ncdot n-fracn(n-1)2=fracn(n+1)2$$
answered Jul 20 at 11:48
gimusi
65.4k73584
(second last inequality) if $n=4$, the second term is $6$, whereas $n(n+1)/2 = 10$?
– Shirish Kulhari
Jul 20 at 11:54
@ShirishKulhari thanks I fix!
– gimusi
Jul 20 at 12:04
add a comment |Â
up vote
0
down vote
For every pair $(i,j)$ with $0leq i<jleq2$ you must add $1$, so the summation gives you the number of such pairs, which is:$$binomn+12=fracn(n+1)2$$
answered Jul 20 at 12:25
Vera
1,706413
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The domain of the integer pairs $(i,j)$ such that
$$0le i<jle n$$ is a triangle inside the square of size $n+1$ by $n+1$, diagonal excluded. Summing the term $1$ means that you just count those elements.
Hence you compute the area of the square minus the diagonal, and halve, giving
$$frac(n+1)^2-(n+1)2=fracn(n+1)2.$$
answered Jul 20 at 12:32
Yves Daoust
111k665204
add a comment |Â
up vote
4
down vote
accepted
The domain of the integer pairs $(i,j)$ such that
$$0le i<jle n$$ is a triangle inside the square of size $n+1$ by $n+1$, diagonal excluded. Summing the term $1$ means that you just count those elements.
Hence you compute the area of the square minus the diagonal, and halve, giving
$$frac(n+1)^2-(n+1)2=fracn(n+1)2.$$
answered Jul 20 at 12:32
Yves Daoust
111k665204
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The domain of the integer pairs $(i,j)$ such that
$$0le i<jle n$$ is a triangle inside the square of size $n+1$ by $n+1$, diagonal excluded. Summing the term $1$ means that you just count those elements.
Hence you compute the area of the square minus the diagonal, and halve, giving
$$frac(n+1)^2-(n+1)2=fracn(n+1)2.$$
answered Jul 20 at 12:32
Yves Daoust
111k665204
The domain of the integer pairs $(i,j)$ such that
$$0le i<jle n$$ is a triangle inside the square of size $n+1$ by $n+1$, diagonal excluded. Summing the term $1$ means that you just count those elements.
Hence you compute the area of the square minus the diagonal, and halve, giving
$$frac(n+1)^2-(n+1)2=fracn(n+1)2.$$
answered Jul 20 at 12:32
Yves Daoust
111k665204
edited Jul 20 at 12:39
answered Jul 20 at 12:32
Yves Daoust
111k665204
answered Jul 20 at 12:32
Yves Daoust
111k665204
answered Jul 20 at 12:32
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
up vote
1
down vote
Note that you can convert the expression into:
$$sum_i=0^n-1bigg(sum_j=i+1^n 1bigg) = sum_i=0^n-1 (n-i) = sum_i=0^n-1 n - sum_i=0^n-1 i = n^2-fracn(n-1)2 = fracn(n+1)2$$
When you see double sums like that, try breaking it up into individual sums. If one index (in this case, $j$) is conditional on another ($i$), the limit(s) of the dependent index can be expressed in terms of the other index. There's no reference/resource as such that comes to my mind, but it should get easier as you solve more and more such problems.
answered Jul 20 at 11:52
Shirish Kulhari
929215
Sir can u provide additional resources regarding this question
– Rafael Nadal
Jul 20 at 11:53
add a comment |Â
up vote
1
down vote
Note that you can convert the expression into:
$$sum_i=0^n-1bigg(sum_j=i+1^n 1bigg) = sum_i=0^n-1 (n-i) = sum_i=0^n-1 n - sum_i=0^n-1 i = n^2-fracn(n-1)2 = fracn(n+1)2$$
When you see double sums like that, try breaking it up into individual sums. If one index (in this case, $j$) is conditional on another ($i$), the limit(s) of the dependent index can be expressed in terms of the other index. There's no reference/resource as such that comes to my mind, but it should get easier as you solve more and more such problems.
answered Jul 20 at 11:52
Shirish Kulhari
929215
Sir can u provide additional resources regarding this question
– Rafael Nadal
Jul 20 at 11:53
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that you can convert the expression into:
$$sum_i=0^n-1bigg(sum_j=i+1^n 1bigg) = sum_i=0^n-1 (n-i) = sum_i=0^n-1 n - sum_i=0^n-1 i = n^2-fracn(n-1)2 = fracn(n+1)2$$
When you see double sums like that, try breaking it up into individual sums. If one index (in this case, $j$) is conditional on another ($i$), the limit(s) of the dependent index can be expressed in terms of the other index. There's no reference/resource as such that comes to my mind, but it should get easier as you solve more and more such problems.
answered Jul 20 at 11:52
Shirish Kulhari
929215
Note that you can convert the expression into:
$$sum_i=0^n-1bigg(sum_j=i+1^n 1bigg) = sum_i=0^n-1 (n-i) = sum_i=0^n-1 n - sum_i=0^n-1 i = n^2-fracn(n-1)2 = fracn(n+1)2$$
When you see double sums like that, try breaking it up into individual sums. If one index (in this case, $j$) is conditional on another ($i$), the limit(s) of the dependent index can be expressed in terms of the other index. There's no reference/resource as such that comes to my mind, but it should get easier as you solve more and more such problems.
answered Jul 20 at 11:52
Shirish Kulhari
929215
edited Jul 20 at 11:57
answered Jul 20 at 11:52
Shirish Kulhari
929215
answered Jul 20 at 11:52
Shirish Kulhari
929215
answered Jul 20 at 11:52
Shirish Kulhari
929215
929215
Sir can u provide additional resources regarding this question
– Rafael Nadal
Jul 20 at 11:53
add a comment |Â
Sir can u provide additional resources regarding this question
– Rafael Nadal
Jul 20 at 11:53
Sir can u provide additional resources regarding this question
– Rafael Nadal
Jul 20 at 11:53
Sir can u provide additional resources regarding this question
– Rafael Nadal
Jul 20 at 11:53
add a comment |Â
up vote
0
down vote
We have that
$$mathopsumsum_0le i<jle n 1=sum_i=0^n-1 left(sum_j=i+1^n 1right)=sum_i=0^n-1 (n-i)=nsum_i=0^n-11-sum_i=0^n-1 i=ncdot n-fracn(n-1)2=fracn(n+1)2$$
answered Jul 20 at 11:48
gimusi
65.4k73584
(second last inequality) if $n=4$, the second term is $6$, whereas $n(n+1)/2 = 10$?
– Shirish Kulhari
Jul 20 at 11:54
@ShirishKulhari thanks I fix!
– gimusi
Jul 20 at 12:04
add a comment |Â
up vote
0
down vote
We have that
$$mathopsumsum_0le i<jle n 1=sum_i=0^n-1 left(sum_j=i+1^n 1right)=sum_i=0^n-1 (n-i)=nsum_i=0^n-11-sum_i=0^n-1 i=ncdot n-fracn(n-1)2=fracn(n+1)2$$
answered Jul 20 at 11:48
gimusi
65.4k73584
(second last inequality) if $n=4$, the second term is $6$, whereas $n(n+1)/2 = 10$?
– Shirish Kulhari
Jul 20 at 11:54
@ShirishKulhari thanks I fix!
– gimusi
Jul 20 at 12:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have that
$$mathopsumsum_0le i<jle n 1=sum_i=0^n-1 left(sum_j=i+1^n 1right)=sum_i=0^n-1 (n-i)=nsum_i=0^n-11-sum_i=0^n-1 i=ncdot n-fracn(n-1)2=fracn(n+1)2$$
answered Jul 20 at 11:48
gimusi
65.4k73584
We have that
$$mathopsumsum_0le i<jle n 1=sum_i=0^n-1 left(sum_j=i+1^n 1right)=sum_i=0^n-1 (n-i)=nsum_i=0^n-11-sum_i=0^n-1 i=ncdot n-fracn(n-1)2=fracn(n+1)2$$
answered Jul 20 at 11:48
gimusi
65.4k73584
edited Jul 20 at 12:03
answered Jul 20 at 11:48
gimusi
65.4k73584
answered Jul 20 at 11:48
gimusi
65.4k73584
answered Jul 20 at 11:48
gimusi
65.4k73584
65.4k73584
(second last inequality) if $n=4$, the second term is $6$, whereas $n(n+1)/2 = 10$?
– Shirish Kulhari
Jul 20 at 11:54
@ShirishKulhari thanks I fix!
– gimusi
Jul 20 at 12:04
add a comment |Â
(second last inequality) if $n=4$, the second term is $6$, whereas $n(n+1)/2 = 10$?
– Shirish Kulhari
Jul 20 at 11:54
@ShirishKulhari thanks I fix!
– gimusi
Jul 20 at 12:04
(second last inequality) if $n=4$, the second term is $6$, whereas $n(n+1)/2 = 10$?
– Shirish Kulhari
Jul 20 at 11:54
(second last inequality) if $n=4$, the second term is $6$, whereas $n(n+1)/2 = 10$?
– Shirish Kulhari
Jul 20 at 11:54
@ShirishKulhari thanks I fix!
– gimusi
Jul 20 at 12:04
@ShirishKulhari thanks I fix!
– gimusi
Jul 20 at 12:04
add a comment |Â
up vote
0
down vote
For every pair $(i,j)$ with $0leq i<jleq2$ you must add $1$, so the summation gives you the number of such pairs, which is:$$binomn+12=fracn(n+1)2$$
answered Jul 20 at 12:25
Vera
1,706413
add a comment |Â
up vote
0
down vote
For every pair $(i,j)$ with $0leq i<jleq2$ you must add $1$, so the summation gives you the number of such pairs, which is:$$binomn+12=fracn(n+1)2$$
answered Jul 20 at 12:25
Vera
1,706413
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For every pair $(i,j)$ with $0leq i<jleq2$ you must add $1$, so the summation gives you the number of such pairs, which is:$$binomn+12=fracn(n+1)2$$
answered Jul 20 at 12:25
Vera
1,706413
For every pair $(i,j)$ with $0leq i<jleq2$ you must add $1$, so the summation gives you the number of such pairs, which is:$$binomn+12=fracn(n+1)2$$
answered Jul 20 at 12:25
Vera
1,706413
answered Jul 20 at 12:25
Vera
1,706413
answered Jul 20 at 12:25
Vera
1,706413
answered Jul 20 at 12:25
Vera
1,706413
1,706413
add a comment |Â
add a comment |Â
1
I am a highschool student who just want explaination please don't close my question
– Rafael Nadal
Jul 20 at 11:42