Evaluating double sums with related indices
I am having trouble finding any techniques that allow people to solve double sums where the indices rely on each other. For example, suppose I have the following sum:
$$sum_{i=1}^{n-1}sum_{j=i+1}^n i + j$$
What techniques can I use to eliminate the sums, and arrive at a simple algebraic expression in terms of only n?
discrete-mathematics summation
asked Nov 24 at 19:53
MegaZeroX
293
add a comment |
I am having trouble finding any techniques that allow people to solve double sums where the indices rely on each other. For example, suppose I have the following sum:
$$sum_{i=1}^{n-1}sum_{j=i+1}^n i + j$$
What techniques can I use to eliminate the sums, and arrive at a simple algebraic expression in terms of only n?
discrete-mathematics summation
asked Nov 24 at 19:53
MegaZeroX
293
Are you looking for a way to find the sum or to a way to simplify the indices?
– gimusi
Nov 24 at 19:58
Ind the end, I want to be able to find the sum. If simplifying the indices helps achieve that goal, then great. I have never learned how to simplify indices in sums. Does that reduce this to only one index variable?
– MegaZeroX
Nov 24 at 20:00
I've just indicated a way to find the sum. There also tecniques to simplify the indices (e.g. double counting) but I can see a clever way now.
– gimusi
Nov 24 at 20:05
add a comment |
I am having trouble finding any techniques that allow people to solve double sums where the indices rely on each other. For example, suppose I have the following sum:
$$sum_{i=1}^{n-1}sum_{j=i+1}^n i + j$$
What techniques can I use to eliminate the sums, and arrive at a simple algebraic expression in terms of only n?
discrete-mathematics summation
asked Nov 24 at 19:53
MegaZeroX
293
I am having trouble finding any techniques that allow people to solve double sums where the indices rely on each other. For example, suppose I have the following sum:
$$sum_{i=1}^{n-1}sum_{j=i+1}^n i + j$$
What techniques can I use to eliminate the sums, and arrive at a simple algebraic expression in terms of only n?
discrete-mathematics summation
discrete-mathematics summation
asked Nov 24 at 19:53
MegaZeroX
293
asked Nov 24 at 19:53
MegaZeroX
293
asked Nov 24 at 19:53
MegaZeroX
293
asked Nov 24 at 19:53
MegaZeroX
293
asked Nov 24 at 19:53
MegaZeroX
293
293
Are you looking for a way to find the sum or to a way to simplify the indices?
– gimusi
Nov 24 at 19:58
Ind the end, I want to be able to find the sum. If simplifying the indices helps achieve that goal, then great. I have never learned how to simplify indices in sums. Does that reduce this to only one index variable?
– MegaZeroX
Nov 24 at 20:00
I've just indicated a way to find the sum. There also tecniques to simplify the indices (e.g. double counting) but I can see a clever way now.
– gimusi
Nov 24 at 20:05
add a comment |
Are you looking for a way to find the sum or to a way to simplify the indices?
– gimusi
Nov 24 at 19:58
Ind the end, I want to be able to find the sum. If simplifying the indices helps achieve that goal, then great. I have never learned how to simplify indices in sums. Does that reduce this to only one index variable?
– MegaZeroX
Nov 24 at 20:00
I've just indicated a way to find the sum. There also tecniques to simplify the indices (e.g. double counting) but I can see a clever way now.
– gimusi
Nov 24 at 20:05
Are you looking for a way to find the sum or to a way to simplify the indices?
– gimusi
Nov 24 at 19:58
Are you looking for a way to find the sum or to a way to simplify the indices?
– gimusi
Nov 24 at 19:58
Ind the end, I want to be able to find the sum. If simplifying the indices helps achieve that goal, then great. I have never learned how to simplify indices in sums. Does that reduce this to only one index variable?
– MegaZeroX
Nov 24 at 20:00
Ind the end, I want to be able to find the sum. If simplifying the indices helps achieve that goal, then great. I have never learned how to simplify indices in sums. Does that reduce this to only one index variable?
– MegaZeroX
Nov 24 at 20:00
I've just indicated a way to find the sum. There also tecniques to simplify the indices (e.g. double counting) but I can see a clever way now.
– gimusi
Nov 24 at 20:05
I've just indicated a way to find the sum. There also tecniques to simplify the indices (e.g. double counting) but I can see a clever way now.
– gimusi
Nov 24 at 20:05
add a comment |
2 Answers
2
active
oldest
votes
We obtain
begin{align*}
color{blue}{sum_{i=1}^{n-1}sum_{j=i+1}^{n}(i+j)}&=sum_{1leq i<jleq n}(i+j)tag{1}\
&=sum_{j=2}^nsum_{i=1}^{j-1}(i+j)tag{2}\
&=sum_{j=2}^nleft(sum_{i=1}^{j-1}i+(j-1)jright)tag{3}\
&=sum_{j=2}^nleft(frac{1}{2}(j-1)j+(j-1)jright)tag{4}\
&=frac{3}{2}sum_{j=1}^n(j-1)jtag{5}\
&=frac{3}{2}left(frac{1}{6}n(n+1)(2n+1)-frac{1}{2}n(n+1)right)tag{6}\
&,,color{blue}{=frac{1}{2}(n-1)n(n+1)}tag{7}
end{align*}
Comment:
In (1) we write the index region somewhat more conveniently.
Note that we assume the index $j$ is a bound variable and so we write $(i+j)$ consequently in parenthesis.
In (2) we exchange the sums so that the inner sum starts now with $i=1$.
In (3) we use the summation formula $sum_{k=1}^n c= ncdot c$.
In (4) we use the summation formula $sum_{k=1}^n k=frac{1}{2}n(n+1)$.
In (5) we do some simplifications and start with $j=1$ without changing anything since we are adding $0$ only.
In (6) we use the formula $sum_{k=1}^nk^2=frac{1}{6}n(n+1)(2n+1)$ and the same as in (4).
In (7) we do some final simplifications.
answered Nov 25 at 16:53
Markus Scheuer
59.8k455143
Thank you for this! I had never hear of bound variables before, and had no idea that simply changing the order of the summation would be so useful! His was very helpful!
– MegaZeroX
Nov 25 at 20:55
@MegaZeroX That's indeed a better way!
– gimusi
Nov 25 at 20:56
@MegaZeroX: You're welcome. If you want to dig deeper into this subject I recommend chapter 2: "Sums" of Concrete Mathematics by D.E. Knuth, R.L. Graham and O. Patashnik.
– Markus Scheuer
Nov 25 at 21:02
1
@MarkusScheuer OK, thanks for the recommendation! I've heard of it, but never had time to give it a shot. I'll hopefully have time for that when summer comes.
– MegaZeroX
Nov 25 at 21:10
add a comment |
We can sum at first the inner summation that is
$$sum_{j=i+1}^n i + j=isum_{j=i+1}^n 1 +sum_{j=i+1}^n j=i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}$$
and then
$$sum_{i=1}^{n-1}sum_{j=i+1}^n i + j=sum_{i=1}^{n-1}left(i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}right)=$$
$$=frac{n(n-1)}{2}sum_{i=1}^{n-1}1+sum_{i=1}^{n-1}left(-frac32i^2+ileft(n-frac12right)right)$$
answered Nov 24 at 20:00
gimusi
1
Ah, OK. I didn't realize that we could simply evaluate the inner sum, substituting i in, and then use that new expression to solve the outer sum. Thanks!
– MegaZeroX
Nov 24 at 20:05
@MegaZeroX Yes that's the standard way to proceed. I don't know if in that case we can also proceed with some clever manipualtion.
– gimusi
Nov 24 at 20:06
add a comment |
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2 Answers
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2 Answers
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We obtain
begin{align*}
color{blue}{sum_{i=1}^{n-1}sum_{j=i+1}^{n}(i+j)}&=sum_{1leq i<jleq n}(i+j)tag{1}\
&=sum_{j=2}^nsum_{i=1}^{j-1}(i+j)tag{2}\
&=sum_{j=2}^nleft(sum_{i=1}^{j-1}i+(j-1)jright)tag{3}\
&=sum_{j=2}^nleft(frac{1}{2}(j-1)j+(j-1)jright)tag{4}\
&=frac{3}{2}sum_{j=1}^n(j-1)jtag{5}\
&=frac{3}{2}left(frac{1}{6}n(n+1)(2n+1)-frac{1}{2}n(n+1)right)tag{6}\
&,,color{blue}{=frac{1}{2}(n-1)n(n+1)}tag{7}
end{align*}
Comment:
In (1) we write the index region somewhat more conveniently.
Note that we assume the index $j$ is a bound variable and so we write $(i+j)$ consequently in parenthesis.
In (2) we exchange the sums so that the inner sum starts now with $i=1$.
In (3) we use the summation formula $sum_{k=1}^n c= ncdot c$.
In (4) we use the summation formula $sum_{k=1}^n k=frac{1}{2}n(n+1)$.
In (5) we do some simplifications and start with $j=1$ without changing anything since we are adding $0$ only.
In (6) we use the formula $sum_{k=1}^nk^2=frac{1}{6}n(n+1)(2n+1)$ and the same as in (4).
In (7) we do some final simplifications.
answered Nov 25 at 16:53
Markus Scheuer
59.8k455143
Thank you for this! I had never hear of bound variables before, and had no idea that simply changing the order of the summation would be so useful! His was very helpful!
– MegaZeroX
Nov 25 at 20:55
@MegaZeroX That's indeed a better way!
– gimusi
Nov 25 at 20:56
@MegaZeroX: You're welcome. If you want to dig deeper into this subject I recommend chapter 2: "Sums" of Concrete Mathematics by D.E. Knuth, R.L. Graham and O. Patashnik.
– Markus Scheuer
Nov 25 at 21:02
1
@MarkusScheuer OK, thanks for the recommendation! I've heard of it, but never had time to give it a shot. I'll hopefully have time for that when summer comes.
– MegaZeroX
Nov 25 at 21:10
add a comment |
We obtain
begin{align*}
color{blue}{sum_{i=1}^{n-1}sum_{j=i+1}^{n}(i+j)}&=sum_{1leq i<jleq n}(i+j)tag{1}\
&=sum_{j=2}^nsum_{i=1}^{j-1}(i+j)tag{2}\
&=sum_{j=2}^nleft(sum_{i=1}^{j-1}i+(j-1)jright)tag{3}\
&=sum_{j=2}^nleft(frac{1}{2}(j-1)j+(j-1)jright)tag{4}\
&=frac{3}{2}sum_{j=1}^n(j-1)jtag{5}\
&=frac{3}{2}left(frac{1}{6}n(n+1)(2n+1)-frac{1}{2}n(n+1)right)tag{6}\
&,,color{blue}{=frac{1}{2}(n-1)n(n+1)}tag{7}
end{align*}
Comment:
In (1) we write the index region somewhat more conveniently.
Note that we assume the index $j$ is a bound variable and so we write $(i+j)$ consequently in parenthesis.
In (2) we exchange the sums so that the inner sum starts now with $i=1$.
In (3) we use the summation formula $sum_{k=1}^n c= ncdot c$.
In (4) we use the summation formula $sum_{k=1}^n k=frac{1}{2}n(n+1)$.
In (5) we do some simplifications and start with $j=1$ without changing anything since we are adding $0$ only.
In (6) we use the formula $sum_{k=1}^nk^2=frac{1}{6}n(n+1)(2n+1)$ and the same as in (4).
In (7) we do some final simplifications.
answered Nov 25 at 16:53
Markus Scheuer
59.8k455143
Thank you for this! I had never hear of bound variables before, and had no idea that simply changing the order of the summation would be so useful! His was very helpful!
– MegaZeroX
Nov 25 at 20:55
@MegaZeroX That's indeed a better way!
– gimusi
Nov 25 at 20:56
@MegaZeroX: You're welcome. If you want to dig deeper into this subject I recommend chapter 2: "Sums" of Concrete Mathematics by D.E. Knuth, R.L. Graham and O. Patashnik.
– Markus Scheuer
Nov 25 at 21:02
1
@MarkusScheuer OK, thanks for the recommendation! I've heard of it, but never had time to give it a shot. I'll hopefully have time for that when summer comes.
– MegaZeroX
Nov 25 at 21:10
add a comment |
We obtain
begin{align*}
color{blue}{sum_{i=1}^{n-1}sum_{j=i+1}^{n}(i+j)}&=sum_{1leq i<jleq n}(i+j)tag{1}\
&=sum_{j=2}^nsum_{i=1}^{j-1}(i+j)tag{2}\
&=sum_{j=2}^nleft(sum_{i=1}^{j-1}i+(j-1)jright)tag{3}\
&=sum_{j=2}^nleft(frac{1}{2}(j-1)j+(j-1)jright)tag{4}\
&=frac{3}{2}sum_{j=1}^n(j-1)jtag{5}\
&=frac{3}{2}left(frac{1}{6}n(n+1)(2n+1)-frac{1}{2}n(n+1)right)tag{6}\
&,,color{blue}{=frac{1}{2}(n-1)n(n+1)}tag{7}
end{align*}
Comment:
In (1) we write the index region somewhat more conveniently.
Note that we assume the index $j$ is a bound variable and so we write $(i+j)$ consequently in parenthesis.
In (2) we exchange the sums so that the inner sum starts now with $i=1$.
In (3) we use the summation formula $sum_{k=1}^n c= ncdot c$.
In (4) we use the summation formula $sum_{k=1}^n k=frac{1}{2}n(n+1)$.
In (5) we do some simplifications and start with $j=1$ without changing anything since we are adding $0$ only.
In (6) we use the formula $sum_{k=1}^nk^2=frac{1}{6}n(n+1)(2n+1)$ and the same as in (4).
In (7) we do some final simplifications.
answered Nov 25 at 16:53
Markus Scheuer
59.8k455143
We obtain
begin{align*}
color{blue}{sum_{i=1}^{n-1}sum_{j=i+1}^{n}(i+j)}&=sum_{1leq i<jleq n}(i+j)tag{1}\
&=sum_{j=2}^nsum_{i=1}^{j-1}(i+j)tag{2}\
&=sum_{j=2}^nleft(sum_{i=1}^{j-1}i+(j-1)jright)tag{3}\
&=sum_{j=2}^nleft(frac{1}{2}(j-1)j+(j-1)jright)tag{4}\
&=frac{3}{2}sum_{j=1}^n(j-1)jtag{5}\
&=frac{3}{2}left(frac{1}{6}n(n+1)(2n+1)-frac{1}{2}n(n+1)right)tag{6}\
&,,color{blue}{=frac{1}{2}(n-1)n(n+1)}tag{7}
end{align*}
Comment:
In (1) we write the index region somewhat more conveniently.
Note that we assume the index $j$ is a bound variable and so we write $(i+j)$ consequently in parenthesis.
In (2) we exchange the sums so that the inner sum starts now with $i=1$.
In (3) we use the summation formula $sum_{k=1}^n c= ncdot c$.
In (4) we use the summation formula $sum_{k=1}^n k=frac{1}{2}n(n+1)$.
In (5) we do some simplifications and start with $j=1$ without changing anything since we are adding $0$ only.
In (6) we use the formula $sum_{k=1}^nk^2=frac{1}{6}n(n+1)(2n+1)$ and the same as in (4).
In (7) we do some final simplifications.
answered Nov 25 at 16:53
Markus Scheuer
59.8k455143
answered Nov 25 at 16:53
Markus Scheuer
59.8k455143
answered Nov 25 at 16:53
Markus Scheuer
59.8k455143
answered Nov 25 at 16:53
Markus Scheuer
59.8k455143
59.8k455143
Thank you for this! I had never hear of bound variables before, and had no idea that simply changing the order of the summation would be so useful! His was very helpful!
– MegaZeroX
Nov 25 at 20:55
@MegaZeroX That's indeed a better way!
– gimusi
Nov 25 at 20:56
@MegaZeroX: You're welcome. If you want to dig deeper into this subject I recommend chapter 2: "Sums" of Concrete Mathematics by D.E. Knuth, R.L. Graham and O. Patashnik.
– Markus Scheuer
Nov 25 at 21:02
1
@MarkusScheuer OK, thanks for the recommendation! I've heard of it, but never had time to give it a shot. I'll hopefully have time for that when summer comes.
– MegaZeroX
Nov 25 at 21:10
add a comment |
Thank you for this! I had never hear of bound variables before, and had no idea that simply changing the order of the summation would be so useful! His was very helpful!
– MegaZeroX
Nov 25 at 20:55
@MegaZeroX That's indeed a better way!
– gimusi
Nov 25 at 20:56
@MegaZeroX: You're welcome. If you want to dig deeper into this subject I recommend chapter 2: "Sums" of Concrete Mathematics by D.E. Knuth, R.L. Graham and O. Patashnik.
– Markus Scheuer
Nov 25 at 21:02
1
@MarkusScheuer OK, thanks for the recommendation! I've heard of it, but never had time to give it a shot. I'll hopefully have time for that when summer comes.
– MegaZeroX
Nov 25 at 21:10
Thank you for this! I had never hear of bound variables before, and had no idea that simply changing the order of the summation would be so useful! His was very helpful!
– MegaZeroX
Nov 25 at 20:55
Thank you for this! I had never hear of bound variables before, and had no idea that simply changing the order of the summation would be so useful! His was very helpful!
– MegaZeroX
Nov 25 at 20:55
@MegaZeroX That's indeed a better way!
– gimusi
Nov 25 at 20:56
@MegaZeroX That's indeed a better way!
– gimusi
Nov 25 at 20:56
@MegaZeroX: You're welcome. If you want to dig deeper into this subject I recommend chapter 2: "Sums" of Concrete Mathematics by D.E. Knuth, R.L. Graham and O. Patashnik.
– Markus Scheuer
Nov 25 at 21:02
@MegaZeroX: You're welcome. If you want to dig deeper into this subject I recommend chapter 2: "Sums" of Concrete Mathematics by D.E. Knuth, R.L. Graham and O. Patashnik.
– Markus Scheuer
Nov 25 at 21:02
1
1
@MarkusScheuer OK, thanks for the recommendation! I've heard of it, but never had time to give it a shot. I'll hopefully have time for that when summer comes.
– MegaZeroX
Nov 25 at 21:10
@MarkusScheuer OK, thanks for the recommendation! I've heard of it, but never had time to give it a shot. I'll hopefully have time for that when summer comes.
– MegaZeroX
Nov 25 at 21:10
add a comment |
We can sum at first the inner summation that is
$$sum_{j=i+1}^n i + j=isum_{j=i+1}^n 1 +sum_{j=i+1}^n j=i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}$$
and then
$$sum_{i=1}^{n-1}sum_{j=i+1}^n i + j=sum_{i=1}^{n-1}left(i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}right)=$$
$$=frac{n(n-1)}{2}sum_{i=1}^{n-1}1+sum_{i=1}^{n-1}left(-frac32i^2+ileft(n-frac12right)right)$$
answered Nov 24 at 20:00
gimusi
1
Ah, OK. I didn't realize that we could simply evaluate the inner sum, substituting i in, and then use that new expression to solve the outer sum. Thanks!
– MegaZeroX
Nov 24 at 20:05
@MegaZeroX Yes that's the standard way to proceed. I don't know if in that case we can also proceed with some clever manipualtion.
– gimusi
Nov 24 at 20:06
add a comment |
We can sum at first the inner summation that is
$$sum_{j=i+1}^n i + j=isum_{j=i+1}^n 1 +sum_{j=i+1}^n j=i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}$$
and then
$$sum_{i=1}^{n-1}sum_{j=i+1}^n i + j=sum_{i=1}^{n-1}left(i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}right)=$$
$$=frac{n(n-1)}{2}sum_{i=1}^{n-1}1+sum_{i=1}^{n-1}left(-frac32i^2+ileft(n-frac12right)right)$$
answered Nov 24 at 20:00
gimusi
1
Ah, OK. I didn't realize that we could simply evaluate the inner sum, substituting i in, and then use that new expression to solve the outer sum. Thanks!
– MegaZeroX
Nov 24 at 20:05
@MegaZeroX Yes that's the standard way to proceed. I don't know if in that case we can also proceed with some clever manipualtion.
– gimusi
Nov 24 at 20:06
add a comment |
We can sum at first the inner summation that is
$$sum_{j=i+1}^n i + j=isum_{j=i+1}^n 1 +sum_{j=i+1}^n j=i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}$$
and then
$$sum_{i=1}^{n-1}sum_{j=i+1}^n i + j=sum_{i=1}^{n-1}left(i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}right)=$$
$$=frac{n(n-1)}{2}sum_{i=1}^{n-1}1+sum_{i=1}^{n-1}left(-frac32i^2+ileft(n-frac12right)right)$$
answered Nov 24 at 20:00
gimusi
1
We can sum at first the inner summation that is
$$sum_{j=i+1}^n i + j=isum_{j=i+1}^n 1 +sum_{j=i+1}^n j=i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}$$
and then
$$sum_{i=1}^{n-1}sum_{j=i+1}^n i + j=sum_{i=1}^{n-1}left(i(n-i)+frac{n(n-1)}{2}-frac{i(i+1)}{2}right)=$$
$$=frac{n(n-1)}{2}sum_{i=1}^{n-1}1+sum_{i=1}^{n-1}left(-frac32i^2+ileft(n-frac12right)right)$$
answered Nov 24 at 20:00
gimusi
1
answered Nov 24 at 20:00
gimusi
1
answered Nov 24 at 20:00
gimusi
1
answered Nov 24 at 20:00
gimusi
1
1
Ah, OK. I didn't realize that we could simply evaluate the inner sum, substituting i in, and then use that new expression to solve the outer sum. Thanks!
– MegaZeroX
Nov 24 at 20:05
@MegaZeroX Yes that's the standard way to proceed. I don't know if in that case we can also proceed with some clever manipualtion.
– gimusi
Nov 24 at 20:06
add a comment |
Ah, OK. I didn't realize that we could simply evaluate the inner sum, substituting i in, and then use that new expression to solve the outer sum. Thanks!
– MegaZeroX
Nov 24 at 20:05
@MegaZeroX Yes that's the standard way to proceed. I don't know if in that case we can also proceed with some clever manipualtion.
– gimusi
Nov 24 at 20:06
Ah, OK. I didn't realize that we could simply evaluate the inner sum, substituting i in, and then use that new expression to solve the outer sum. Thanks!
– MegaZeroX
Nov 24 at 20:05
Ah, OK. I didn't realize that we could simply evaluate the inner sum, substituting i in, and then use that new expression to solve the outer sum. Thanks!
– MegaZeroX
Nov 24 at 20:05
@MegaZeroX Yes that's the standard way to proceed. I don't know if in that case we can also proceed with some clever manipualtion.
– gimusi
Nov 24 at 20:06
@MegaZeroX Yes that's the standard way to proceed. I don't know if in that case we can also proceed with some clever manipualtion.
– gimusi
Nov 24 at 20:06
add a comment |
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Are you looking for a way to find the sum or to a way to simplify the indices?
– gimusi
Nov 24 at 19:58
Ind the end, I want to be able to find the sum. If simplifying the indices helps achieve that goal, then great. I have never learned how to simplify indices in sums. Does that reduce this to only one index variable?
– MegaZeroX
Nov 24 at 20:00
I've just indicated a way to find the sum. There also tecniques to simplify the indices (e.g. double counting) but I can see a clever way now.
– gimusi
Nov 24 at 20:05