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Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(N+2)/6.
Version: 2
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Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.
Version: 3
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Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.

Retroactive class chatter

      Somebody asks if "summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?


...okay, as much as he doesn't like stopping to prove everything, that way of looking at things does somewhat alarm his dath ilani sensibilities.

Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2].  Proof by generalizing examples:

sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.

Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly dubious reasoning that will create a good intuition for what's really going on.  So instead he'll just quickly write on the board that: 

([N+1])([N+1]+1)(2[N+1]+1)/6  - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1)  *  [(N+2)(2N + 3)  -  (N)(2N+1)]  /  6
= (N+1)  *  [2N^2 + 7N + 6 - 2N^2 - N]  /  6
= (N+1)  *  [6N + 6]  /  6
= (N+1)^2


=> let  f(N) = N(N+1)(2N+1)/6  in
f(N) + (N+1)^2 = f(N+1)
also f(0)=0 just to be clear
Version: 4
Fields Changed Content
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Content
Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.

Retroactive class chatter

      Somebody asks if "summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?


...okay, as much as he doesn't like stopping to prove everything, that way of looking at things does somewhat alarm his dath ilani sensibilities.

Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2].  Proof by generalizing examples:

sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.

Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly dubious reasoning that will create a good intuition for what's really going on.  So instead he'll just quickly write on the board that:

let   f(N) = N(N+1)(2N+1)/6

in    f(0) = 0
      f(N+1) = f(N) + (N+1)^2

proof:

  0(0+1)(2(0)+1
= 0(1)(3)
= 0

  f(N+1) - f(N)
= ([N+1])([N+1]+1)(2[N+1]+1)/6  - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1)  *  [(N+2)(2N + 3)  -  (N)(2N+1)]  /  6
= (N+1)  *  [2N^2 + 7N + 6 - 2N^2 - N]  /  6
= (N+1)  *  [6N + 6]  /  6
= (N+1)^2
Version: 5
Fields Changed Content
Updated
Content
Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.

Retroactive class chatter

      Somebody asks if "summing 1 to N gives N(N +1)/2, and summing squares gives N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?


...this question does somewhat alarm Keltham's dath ilani sensibilities.

Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2].  Proof by generalizing examples:

sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.

Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly quick and dubious reasoning that will create a good intuition for what's really going on.  So instead he'll just quickly write on the board that:

let   f(N) = N(N+1)(2N+1)/6

in    f(0) = 0
      f(N+1) = f(N) + (N+1)^2

proof:

  0(0+1)(2(0)+1
= 0(1)(3)
= 0

  f(N+1) - f(N)
= ([N+1])([N+1]+1)(2[N+1]+1)/6  - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1)  *  [(N+2)(2N + 3)  -  (N)(2N+1)]  /  6
= (N+1)  *  [2N^2 + 7N + 6 - 2N^2 - N]  /  6
= (N+1)  *  [6N + 6]  /  6
= (N+1)^2
Version: 6
Fields Changed Content
Updated
Content
Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.

Retroactive class chatter

      Somebody asks if "summing 1 to N gives N(N +1)/2, and summing squares gives N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?


...

...Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2].  Proof by generalizing examples:

sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.

Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly quick and dubious reasoning that will create a good intuition for what's really going on.  So instead he'll just quickly write on the board that:

let   f(N) = N(N+1)(2N+1)/6

in    f(0) = 0
      f(N+1) = f(N) + (N+1)^2

proof:

  0(0+1)(2(0)+1
= 0(1)(3)
= 0

  f(N+1) - f(N)
= ([N+1])([N+1]+1)(2[N+1]+1)/6  - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1)  *  [(N+2)(2N + 3)  -  (N)(2N+1)]  /  6
= (N+1)  *  [2N^2 + 7N + 6 - 2N^2 - N]  /  6
= (N+1)  *  [6N + 6]  /  6
= (N+1)^2
Version: 7
Fields Changed Content
Updated
Content
Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.

Retroactive class chatter

      Somebody asks if "summing 1 to N gives N(N +1)/2, and summing squares gives N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?


...

...Keltham is not usually one to duck into details that aren't relative to the main line of argument, but this question is sort of alarming.

Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2].  Proof by generalizing examples:

   sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.

Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly quick and dubious reasoning that will create a good intuition for what's really going on.  So instead he'll just quickly write on the board that:

let   f(N) = N(N+1)(2N+1)/6

in    f(0) = 0
      f(N+1) = f(N) + (N+1)^2

proof:

  0(0+1)(2(0)+1
= 0(1)(3)
= 0

  f(N+1) - f(N)
= ([N+1])([N+1]+1)(2[N+1]+1)/6  - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1)  *  [(N+2)(2N + 3)  -  (N)(2N+1)]  /  6
= (N+1)  *  [2N^2 + 7N + 6 - 2N^2 - N]  /  6
= (N+1)  *  [6N + 6]  /  6
= (N+1)^2

If they're not happy with this logical proof by certain reasoning because, like, why, they should consider visualizing stacking a bunch of n-sided squares on top of each other, to form a figure that pretty reasonably looks like it should take up about a third of a cube of size N.
Version: 8
Fields Changed Content
Updated
Content
Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.

Retroactive class chatter

      Somebody asks if "summing 1 to N gives N(N +1)/2, and summing squares gives N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?


...

...Keltham is not usually one to duck into details that aren't relative to the main line of argument, but this question is sort of alarming.

Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2].  Proof by generalizing examples:

   sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.

Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly quick and dubious reasoning that will create a good intuition for what's really going on.  So instead he'll just quickly write on the board that:

let   f(N) = N(N+1)(2N+1)/6

in    f(0) = 0
      f(N+1) = f(N) + (N+1)^2

proof:

  0(0+1)(2(0)+1
= 0(1)(3)
= 0

  f(N+1) - f(N)
= ([N+1])([N+1]+1)(2[N+1]+1)/6  - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1)  *  [(N+2)(2N + 3)  -  (N)(2N+1)]  /  6
= (N+1)  *  [2N^2 + 7N + 6 - 2N^2 - N]  /  6
= (N+1)  *  [6N + 6]  /  6
= (N+1)^2

If they're not happy with this logical proof by certain reasoning because, like, why, they should consider visualizing stacking a bunch of n-sided squares on top of each other, to form a figure that pretty reasonably looks like it should take up about a third of a cube of size N.

Now, back to the problem of evaluating:

[1^2 + 2^2 + 3^2 + 4^2 + ...] / ( [1 + 2 + 3 + 4 + ...] times infinity )
Version: 9
Fields Changed Content
Updated
Content
Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.

Retroactive class chatter

      Somebody asks if "summing 1 to N gives N(N +1)/2, and summing squares gives N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?


...

...Keltham is not usually one to duck into details that aren't relative to the main line of argument, but this question is sort of alarming.

Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2].  Proof by generalizing examples:

   sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.

Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly hasty intuitive reasoning that will create a good sense for what's really going on.  So instead he'll just quickly write on the board that:

let   f(N) = N(N+1)(2N+1)/6

in    f(0) = 0
      f(N+1) = f(N) + (N+1)^2

proof:

  0(0+1)(2(0)+1
= 0(1)(3)
= 0

  f(N+1) - f(N)
= ([N+1])([N+1]+1)(2[N+1]+1)/6  - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1)  *  [(N+2)(2N + 3)  -  (N)(2N+1)]  /  6
= (N+1)  *  [2N^2 + 7N + 6 - 2N^2 - N]  /  6
= (N+1)  *  [6N + 6]  /  6
= (N+1)^2

If they're not happy with this logical proof by certain reasoning - because, like, why - they should consider visualizing stacking a bunch of n-sided squares on top of each other, to form a figure that pretty much looks like it should take up about a third of a cube of size N.

Now, back to the problem of evaluating:

[1^2 + 2^2 + 3^2 + 4^2 + ...] / { [1 + 2 + 3 + 4 + ...] times infinity }
Version: 10
Fields Changed Content
Updated
Content
Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.

Parenthetical retroactive class chatter

      Somebody asks if "summing 1 to N gives N(N +1)/2, and summing squares gives N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?


...

...Keltham is not usually one to duck into details that aren't relative to the main line of argument, but this question is sort of alarming.

Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2].  Proof by generalizing examples:

   sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.

Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly hasty intuitive reasoning that will create a good sense for what's really going on.  So instead he'll just quickly write on the board that:

let   f(N) = N(N+1)(2N+1)/6

in    f(0) = 0
      f(N+1) = f(N) + (N+1)^2

proof:

  0(0+1)(2(0)+1
= 0(1)(3)
= 0

  f(N+1) - f(N)
= ([N+1])([N+1]+1)(2[N+1]+1)/6  - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1)  *  [(N+2)(2N + 3)  -  (N)(2N+1)]  /  6
= (N+1)  *  [2N^2 + 7N + 6 - 2N^2 - N]  /  6
= (N+1)  *  [6N + 6]  /  6
= (N+1)^2

If they're not happy with this logical proof by certain reasoning - because, like, why - they should consider visualizing stacking a bunch of n-sided squares on top of each other, to form a figure that pretty much looks like it should take up about a third of a cube of size N.

Now, back to the problem of evaluating:

[1^2 + 2^2 + 3^2 + 4^2 + ...] / { [1 + 2 + 3 + 4 + ...] times infinity }
Version: 11
Fields Changed Content, editor mode
Updated
Content
Iarwain

...one of the older students in the background does happen to know the useful fact that summing from 1 to N gives you N(N +1)/2, and summing the squares from 1 to N gives you N(N+1)(2N+1)/6.

Parenthetical retroactive class chatter

      Somebody asks if "summing 1 to N gives N(N +1)/2, and summing squares gives N(N+1)(2N+1)/6" are to be treated as brute facts or the sort of things you can prove?


...

...Keltham is not usually one to duck into details that aren't relative to the main line of argument, but this question is sort of alarming.

Keltham will observe that [1 + 2 + 3 + 4 + 5 + 6] = [(1 + 6) + (2 + 5) + (3 + 4)] = [7 + 7 + 7], and that [1 + 2 + 3 + 4 + 5] = [(1 + 5) + (2 + 4) + 3] = [6 + 6 + 6/2].  Proof by generalizing examples:

   sum of [1 + 2 + ... + N] with N terms
= half that many [(N + 1)] terms, so N/2 * (N+1)
= N(N + 1) / 2.

Keltham doesn't really want to stop and prove the sum of n^2 from 1 to N is N(N+1)(2N+1)/6 using properly hasty intuitive reasoning that will create a good sense for what's really going on.  So instead he'll just quickly write on the board that:

let   f(N) = N(N+1)(2N+1)/6

in    f(0) = 0
      f(N+1) = f(N) + (N+1)^2

proof:

  0(0+1)(2(0)+1)
= 0(1)(3)
= 0

  f(N+1) - f(N)
= ([N+1])([N+1]+1)(2[N+1]+1)/6  - N(N+1)(2N+1)/6
= (N+1)(N+2)(2N + 3)/6 - N(N+1)(2N+1)/6
= (N+1)  *  [(N+2)(2N + 3)  -  (N)(2N+1)]  /  6
= (N+1)  *  [2N^2 + 7N + 6 - 2N^2 - N]  /  6
= (N+1)  *  [6N + 6]  /  6
= (N+1)^2

If they're not happy with this logical proof by certain reasoning - because, like, why - they should consider visualizing stacking a bunch of n-sided squares on top of each other, to form a figure that pretty much looks like it should take up about a third of a cube of size N.

Now, back to the problem of evaluating:

[1^2 + 2^2 + 3^2 + 4^2 + ...] / { [1 + 2 + 3 + 4 + ...] times infinity }