[OT] Two Oh Two Five - a breakdown

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

Post by occam
Last one was 1936, the next one
will be the year 2116.
Don't all rush to your numerologist, they are all charlatans.

--
Athel -- French and British, living in Marseilles for 37 years; mainly
in England until 1987.

Anders D. Nygaard

2025-01-02 20:04:34 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

Making it an instance of a general result:

(1+2+...+n)^2 = 1^3+2^3+...n^3

Post by occam
Last one was 1936, the next one
will be the year 2116.
Don't all rush to your numerologist, they are all charlatans.

Anders, Denmark

jerryfriedman

2025-01-02 20:26:44 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

..

Thanks, if I ever knew that, I forgot it. My math
students might hear it this term and next.

--
Jerry Friedman

--

Athel Cornish-Bowden

2025-01-02 21:11:16 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Yes. I wondered about that, but was too lazy to look more deeply.

Post by occam
Last one was 1936, the next one
will be the year 2116.
Don't all rush to your numerologist, they are all charlatans.

Anders, Denmark

--
Athel -- French and British, living in Marseilles for 37 years; mainly
in England until 1987.

occam

2025-01-03 08:27:49 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

Peter Moylan

2025-01-03 09:49:02 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

--
Peter Moylan ***@pmoylan.org http://www.pmoylan.org
Newcastle, NSW

Peter Moylan

2025-01-03 10:07:09 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Checked! Yes, it's easy. Start with the well-known fact that
sum(1..N) = N(N+1)/2
Now you want to prove that
(sum (1..N) j)^2 = sum(1..n) j^3
Assume that it's true for the N-1 case. Expand out the quadratic on the
left as
(sum(1..N) j )^2 = (sum(1..N-1) j )^2 + 2N sum(1..N-1) j + N^2
which, by the inductive assumption, is
= (sum(1..N-1) j^3 + N^2 (N - 1) + N^2
and Bob's your mother's brother.

--
Peter Moylan ***@pmoylan.org http://www.pmoylan.org
Newcastle, NSW

Hibou

2025-01-03 10:10:16 UTC

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Checked! Yes, it's easy. Start with the well-known fact that
       sum(1..N) = N(N+1)/2
Now you want to prove that
      (sum (1..N) j)^2 = sum(1..n) j^3
Assume that it's true for the N-1 case. Expand out the quadratic on the
left as
      (sum(1..N) j )^2 = (sum(1..N-1) j )^2 + 2N sum(1..N-1) j + N^2
which, by the inductive assumption, is
                               = (sum(1..N-1) j^3 + N^2 (N - 1) + N^2
and Bob's your mother's brother.

Bravo!

occam

2025-01-03 12:31:33 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Checked! Yes, it's easy. Start with the well-known fact that
       sum(1..N) = N(N+1)/2
Now you want to prove that
      (sum (1..N) j)^2 = sum(1..n) j^3
Assume that it's true for the N-1 case. Expand out the quadratic on the
left as
      (sum(1..N) j )^2 = (sum(1..N-1) j )^2 + 2N sum(1..N-1) j + N^2
which, by the inductive assumption, is
                               = (sum(1..N-1) j^3 + N^2 (N - 1) + N^2
and Bob's your mother's brother.

Thanks. Proof by induction did cross my mind. However the coffee on
induction cooker took precedence. Will have a go myself working out the
details.

jerryfriedman

2025-01-03 14:02:58 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Checked! Yes, it's easy. Start with the well-known fact that
sum(1..N) = N(N+1)/2
Now you want to prove that
(sum (1..N) j)^2 = sum(1..n) j^3
Assume that it's true for the N-1 case.

Interesting. I was taught to assume it's true for N
and show that it follows for N+1. Same thing, of
course.

Post by Peter Moylan
Expand out the quadratic on the
left as
(sum(1..N) j )^2 = (sum(1..N-1) j )^2 + 2N sum(1..N-1) j + N^2
which, by the inductive assumption, is
= (sum(1..N-1) j^3 + N^2 (N - 1) + N^2
and Bob's your mother's brother.

And you get your choice of distributing or factoring
out the N^2.

For those who don't like proofs by induction, other
proofs are available on line.

--
Jerry Friedman

--

occam

2025-01-03 14:42:12 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/=
z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Checked! Yes, it's easy. Start with the well-known fact that
sum(1..N) = N(N+1)/2
Now you want to prove that
(sum (1..N) j)^2 = sum(1..n) j^3
Assume that it's true for the N-1 case.

Interesting. I was taught to assume it's true for N
and show that it follows for N+1. Same thing, of
course.

And you get your choice of distributing or factoring
out the N^2.
For those who don't like proofs by induction, other
proofs are available on line.

I like them. There is an element of a domino effect to them.
(An exercise to set your students next semester.)

jerryfriedman

2025-01-03 17:43:00 UTC

Post by Peter Moylan
Checked! Yes, it's easy. Start with the well-known fact that
sum(1..N) = N(N+1)/2
Now you want to prove that
(sum (1..N) j)^2 = sum(1..n) j^3

[etc.]

Post by jerryfriedman
For those who don't like proofs by induction, other
proofs are available on line.

I like them. There is an element of a domino effect to them.
(An exercise to set your students next semester.)

Yes, the domino effect is fun. The objection I've
heard to proofs by induction is that they don't give
you insight into why something is true, but for
a finite series like this, it shows you that
extending the linear series adds (n+1)^3 to the
total, which gives some satisfying understanding.

--
Jerry Friedman

--

J. J. Lodder

2025-01-11 08:33:59 UTC

Post by jerryfriedman
..

Post by Peter Moylan
Checked! Yes, it's easy. Start with the well-known fact that
sum(1..N) = N(N+1)/2
Now you want to prove that
(sum (1..N) j)^2 = sum(1..n) j^3

[etc.]

Post by jerryfriedman
For those who don't like proofs by induction, other
proofs are available on line.

I like them. There is an element of a domino effect to them.
(An exercise to set your students next semester.)

For a perhaps more satisfying? understanding you can see
<https://mathmunch.org/2016/09/29/the-dice-lab-sum-of-cubes-and-double-polyhedra/>

Jan

Peter Moylan

2025-01-03 22:56:46 UTC

Post by Peter Moylan
I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Checked! Yes, it's easy. Start with the well-known fact that
sum(1..N) = N(N+1)/2
Now you want to prove that
(sum (1..N) j)^2 = sum(1..n) j^3
Assume that it's true for the N-1 case.

Interesting. I was taught to assume it's true for N
and show that it follows for N+1. Same thing, of
course.

Because it's the same thing, I sometimes take one approach and sometimes
the other. I don't have any strong leaning either way.

The difference does matter in the classic proof that all horses have the
same colour.

--
Peter Moylan ***@pmoylan.org http://www.pmoylan.org
Newcastle, NSW

J. J. Lodder

2025-01-03 13:38:20 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Of course, wihout pen and paper even,
if you hapen to remember the sum of an arithmetic series,
or if you derive it on the spot,

Jan

occam

2025-01-06 10:50:18 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Of course, wihout pen and paper even,
if you hapen to remember the sum of an arithmetic series,
or if you derive it on the spot,

Bah! Idle bragging. I /do/ happen to know the sum of the arithmetic
series as N*(N+1)/2 .

However, I'll be buggered if I can square that in my head AND subtract
it from the square of (N+1)(N+2)/2 . (The result should be (N+1)^3 ).

If you can do all that in your head, you should not be wasting your time
here with mere mortals.

jerryfriedman

2025-01-06 15:05:37 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Of course, wihout pen and paper even,
if you hapen to remember the sum of an arithmetic series,
or if you derive it on the spot,

Bah! Idle bragging. I /do/ happen to know the sum of the arithmetic
series as N*(N+1)/2 .
However, I'll be buggered if I can square that in my head AND subtract
it from the square of (N+1)(N+2)/2 . (The result should be (N+1)^3 ).
If you can do all that in your head, you should not be wasting your time
here with mere mortals.

You don't have to square it in your head. What you
have to do (unless you're von Neumann) is realize
you don't have to square it in your head. See Peter's
proof.

--
Jerry Friedman

--

J. J. Lodder

2025-01-06 21:55:37 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Of course, wihout pen and paper even,
if you hapen to remember the sum of an arithmetic series,
or if you derive it on the spot,

Bah! Idle bragging. I /do/ happen to know the sum of the arithmetic
series as N*(N+1)/2 .
However, I'll be buggered if I can square that in my head AND subtract
it from the square of (N+1)(N+2)/2 . (The result should be (N+1)^3 ).
If you can do all that in your head, you should not be wasting your time
here with mere mortals.

You don't have to square it in your head. What you
have to do (unless you're von Neumann) is realize
you don't have to square it in your head. See Peter's
proof.

Mwah, Von Neumann probably would not have to calculate at all.
He would have seen it immediately as some trivial (to him)
corrolary of some obscure property of the zeta-function,

Jan
(veux pas le savoir)

J. J. Lodder

2025-01-06 20:02:31 UTC

A friend in Madrid sent us this yesterday.

Post by occam
Also 2025 is a square number = 45^2.

In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

Bloody marvellous! Thanks. Is there a formal proof for all n (which is
simpler than, say the proof for Fermat's Theorem x^n + y^n =/= z^n ) ?

I don't have a pen in my hand just now, but there should be a simple
proof by induction.

Of course, wihout pen and paper even,
if you hapen to remember the sum of an arithmetic series,
or if you derive it on the spot,

Bah! Idle bragging.

Not really.

Post by occam
I /do/ happen to know the sum of the arithmetic series as N*(N+1)/2 .
However, I'll be buggered if I can square that in my head AND subtract
it from the square of (N+1)(N+2)/2 . (The result should be (N+1)^3 ).

You shouldn't square.

Post by occam
If you can do all that in your head, you should not be wasting your time
here with mere mortals.

What's wrong with wasting time?

Jan

Bertel Lund Hansen

2025-01-03 08:52:51 UTC

Post by Athel Cornish-Bowden
In other words (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 2025.

(1+2+...+n)^2 = 1^3+2^3+...n^3

I didn't know that. Is the proof uncomplicated?

--
Bertel
Kolt, Denmark

Snidely

2025-01-02 17:54:36 UTC

On Thursday or thereabouts, occam asked ...

Post by occam
OK, not exactly the stuff of AUE, however I came across this numerical
breakdown of the number '2025'.
1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 = 2025 (the sum
of the cubes of 1, 2, 3,...,9)
Also 2025 is a square number = 45^2. Last one was 1936, the next one
will be the year 2116.
Don't all rush to your numerologist, they are all charlatans.

I was statisfied with it being int(str(4*5) + str(5*5))

/dps

--
Ieri, oggi, domani

Aidan Kehoe

2025-01-03 06:35:35 UTC

Post by Snidely
On Thursday or thereabouts, occam asked ...

I was statisfied with it being int(str(4*5) + str(5*5))

Is that Python?

What does *your* numerologist say?

--
‘As I sat looking up at the Guinness ad, I could never figure out /
How your man stayed up on the surfboard after fourteen pints of stout’
(C. Moore)

Hibou

2025-01-03 09:40:51 UTC

Post by Aidan Kehoe

Post by Snidely
On Thursday or thereabouts, occam asked ...

I was statisfied with it being int(str(4*5) + str(5*5))

Is that Python?
What does *your* numerologist say?

42.

Peter Moylan

2025-01-03 22:48:05 UTC

Post by Snidely
On Thursday or thereabouts, occam asked ...

I was statisfied with it being int(str(4*5) + str(5*5))

My son has just mentioned
(20 + 25) (20 + 25) = 2025

--
Peter Moylan ***@pmoylan.org http://www.pmoylan.org
Newcastle, NSW

lar3ryca

2025-01-04 06:39:18 UTC

Fascinating. When I read this, I wondered if bc (the Linux command line
basic calculator) would recognize the superscripts as powers.
Sure enough, when I pasted it into the terminal, it showed it as

1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3

And gave me the right answer.

Post by occam
Also 2025 is a square number = 45^2. Last one was 1936, the next one
will be the year 2116.
Don't all rush to your numerologist, they are all charlatans.

--
If you jog in a jogging suit, lounge in lounging pyjamas, and smoke in
a smoking jacket, WHY would anyone want to wear a windbreaker??

lar3ryca

2025-01-04 06:52:14 UTC

Post by lar3ryca

Fascinating. When I read this, I wondered if bc (the Linux command line
basic calculator) would recognize the superscripts as powers.
Sure enough, when I pasted it into the terminal, it showed it as
1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3

Weird!

What Linux actually showed in the terminal when I pasted that, was

1 caret 3 ...

but when I read it from Eternal September, it had changed it back to
1 superscript 3 ...

and just now, when I am composing this response, it has changed the
quoted part back to

1 superscript 3 ...

Post by lar3ryca
And gave me the right answer.

Post by occam
Also 2025 is a square number = 45^2. Last one was 1936, the next one
will be the year 2116.
Don't all rush to your numerologist, they are all charlatans.

--
If you jog in a jogging suit, lounge in lounging pyjamas, and smoke in
a smoking jacket, WHY would anyone want to wear a windbreaker??

Snidely

2025-01-04 10:38:12 UTC