Excalibur Identity

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Excalibur Identity

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130 posts

#1 h Apr 3, 2025, 3:27 PM

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proof is below

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#2 h Apr 3, 2025, 3:58 PM • 1 Y

Y by franklin2013

yes, it is well known that the sum of the first $k$ odd numbers is $k^2$ . perhaps the simplest proof of this is by induction

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GallopingUnicorn45

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GallopingUnicorn45

#3 h Apr 3, 2025, 4:07 PM

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bro excalibur is a las vegas hotel lol

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#4 h Apr 3, 2025, 5:07 PM

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Wait how did you exactly prove it? Like I don't really understand what do you mean by proving through patterns? Did you assume that "yeah it works for the first few terms so it must work for all the terms"?

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130 posts

#5 h Apr 3, 2025, 10:55 PM

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no difference of powers I believe

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130 posts

#6 h Apr 3, 2025, 11:01 PM

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SomeonecoolLovesMaths wrote:

Wait how did you exactly prove it? Like I don't really understand what do you mean by proving through patterns? Did you assume that "yeah it works for the first few terms so it must work for all the terms"?

no the pattern was the difference between the squares

x^2 - (x-1)^2 = (x-(x-1))(x+(x+1)) which further simplifies to 2x-1

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#7 h Apr 3, 2025, 11:28 PM

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for all x

x^2 = (2(1) - 1) + (2(2) - 1) ... + 2(x) - 1

by summation

it's what I wrote but x>= 0

starts from 1 because 2n-1 = difference and smallest is 1^2 - 0 = 1

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mathprodigy2011

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mathprodigy2011

#8 h Apr 4, 2025, 1:58 AM

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SomeonecoolLovesMaths wrote:

Wait how did you exactly prove it? Like I don't really understand what do you mean by proving through patterns? Did you assume that "yeah it works for the first few terms so it must work for all the terms"?

basically, you assume it works for n and then with that you prove it works for (n+1). This is enough to prove that it works for all integers as long as you prove the first case or n=1. Then by your proof; it should work for n=2,3,4,5, etc.

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#9 h Apr 4, 2025, 2:08 AM

Y by

mathprodigy2011 wrote:

SomeonecoolLovesMaths wrote:

Wait how did you exactly prove it? Like I don't really understand what do you mean by proving through patterns? Did you assume that "yeah it works for the first few terms so it must work for all the terms"?

basically, you assume it works for n and then with that you prove it works for (n+1). This is enough to prove that it works for all integers as long as you prove the first case or n=1. Then by your proof; it should work for n=2,3,4,5, etc.

(aka mathematical induction, or just induction)

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#10 h Apr 4, 2025, 12:21 PM

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For a different proof, we can take the pairs (1,2n-1), (3,2n-3), ... If $n$ is even, there are $\tfrac{n}{2}$ pairs, so the sum of them is $\tfrac{n}{2}(2n-k+k)=n^2$ If odd, then there are $\tfrac{n-1}{2}$ pairs and there is $n$ in the middle, which sum to $2n(\tfrac{n-1}{2})+n=n^2.$

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#11 h Apr 6, 2025, 12:40 PM

Y by

mathprodigy2011 wrote:

SomeonecoolLovesMaths wrote:

Wait how did you exactly prove it? Like I don't really understand what do you mean by proving through patterns? Did you assume that "yeah it works for the first few terms so it must work for all the terms"?

basically, you assume it works for n and then with that you prove it works for (n+1). This is enough to prove that it works for all integers as long as you prove the first case or n=1. Then by your proof; it should work for n=2,3,4,5, etc.

Yeah you can of course do it using Induction. I was asking him if he had done it that way or just assumed that it would work looking at the patterns itself.

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mathprodigy2011

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mathprodigy2011

#12 h Apr 6, 2025, 2:49 PM

Y by

SomeonecoolLovesMaths wrote:

mathprodigy2011 wrote:

SomeonecoolLovesMaths wrote:

Wait how did you exactly prove it? Like I don't really understand what do you mean by proving through patterns? Did you assume that "yeah it works for the first few terms so it must work for all the terms"?

basically, you assume it works for n and then with that you prove it works for (n+1). This is enough to prove that it works for all integers as long as you prove the first case or n=1. Then by your proof; it should work for n=2,3,4,5, etc.

Yeah you can of course do it using Induction. I was asking him if he had done it that way or just assumed that it would work looking at the patterns itself.

oh my bad

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#13 h Apr 6, 2025, 2:52 PM

Y by

mathprodigy2011 wrote:

SomeonecoolLovesMaths wrote:

mathprodigy2011 wrote:

basically, you assume it works for n and then with that you prove it works for (n+1). This is enough to prove that it works for all integers as long as you prove the first case or n=1. Then by your proof; it should work for n=2,3,4,5, etc.

Yeah you can of course do it using Induction. I was asking him if he had done it that way or just assumed that it would work looking at the patterns itself.

oh my bad

No problem. Just a misunderstanding perhaps.

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477 posts

#14 h Apr 6, 2025, 8:52 PM

Y by

For yet another proof:
$\sum_{k=1}^{n} (2k-1)$ $=2 \sum_{k=1}^n k - \sum_{k=1}^n 1$ $=n(n+1) - n = n^2.$

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