Excalibur Identity

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jjsunpu

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jjsunpu

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

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proof is below

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Reason: idk

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HockeyMaster85

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HockeyMaster85

#2 h Apr 3, 2025, 3:58 PM • 1 Y

Y by franklin2013

yes, it is well know 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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jjsunpu

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jjsunpu

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

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no difference of powers I believe

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jjsunpu

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jjsunpu

#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

This post has been edited 1 time. Last edited by jjsunpu, Apr 3, 2025, 11:01 PM
Reason: someone teach me code

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jjsunpu

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jjsunpu

#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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MineCuber

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MineCuber

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

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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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fruitmonster97

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fruitmonster97

#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 Yesterday at 12:40 PM

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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 Yesterday at 2:49 PM

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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 Yesterday at 2:52 PM

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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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anduran

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anduran

#14 h Yesterday at 8:52 PM

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