Why is a squared root of sum of cubes equal to a sum of the numbers?
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- Опубликовано: 8 сен 2024
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In this video I show the relationship between the sum of the numbers and the sum of their cubes.
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Nicomachus's theorem!
Thanks! Didn’t know it had a name)
nice
Exactly right
the answer is 2049300 by the way (simple math)
@@inbarreuven4111yeah, simple
(N)(N+1)/2
N=2024
1012*2025 is so simple a 4th grader can do it
“How would you approach this problem?”
“Ah me? Oh, I wouldn’t.”
Thanks for the comment)
Then we are friends 🤝😂
😂😂😂😂😂 youre funny
I would probably try but still fail miserably and give up in the middle of it
Lmao😂
Sigma n³=[n(n+1)/2]²
So square root makes it n(n+1)/2 which is sigma n
Exactly!
Sick!!!!!🥳
Ahhhh true
You mean Sigma n³=Square of Sigma n
@@888_kaiwalyarangle6 Yes sigma n = ((n(n+1))/2 and sigma n^3= [((n(n+1))/2]^2
S = [n²(n + 1)²]/4.... Take square root of that, we get S= [n(n+1)]/2, which is sum of the first n natural numbers, thats why it works
Proof by mathematical induction.
Proposition: sum(n^3)=(sum(n))^2
It's valid when n=1
It's also valid for n=n+1 (increment is (n+1)^3 for both slides)
QED
Thanks for the hint.
nah , he is asking why it works , but induction can't show the rationale for it
I think it should be this
the default here is Σn mean summation of x from x=1 to x=n . Just for simpler illustration.
Δ(Σn)²
=[Σ(n+1)]²-[Σn]²
=(Σ(n+1)-Σ(n))(Σ(n+1)+Σ(n))
=(n+1)(2Σ(n)+n+1)
=(n+1)(n(n+1)+(n+1)) {useΣn=0.5n(n+1)}
=(n+1)(n+1)(n+1)
=(n+1)³
now forget the above default
so ΣΔ[Σ(n)]²=Σ(n+1)³
(Σ(n+1))²-(Σ(0))²=Σ(n+1)³
(Σn)²=Σn³
@@hiyayahiyaya5645where did the 0 come from
@@hiyayahiyaya5645a proof is a proof, induction is a perfectly good answer
Proof by induction is how I learnt to prove it, but I would love to see how the formula was originally made.
Simple pattern recognition, often times the most straightforward answer is the correct one which is why induction is so helpful to verify these claims
Some of cube of n Natural Numbers is [n(n+1)/2]²...
The answer lies in arithmetic progressions
That is true however just looking at a pattern for the first few terms you see isn't proving it works for any numbers out. The way you would know it always works is because the sum of cubes of n natural numbers is ( n^2 * (n + 1)^2 ) / 4, taking the square root of this brings n * ( n + 1 ) / 2 which is the sum of consecutive natural numbers up to n.
Can you explain this more
@noname-ed2un There are 2 formulas we need to know. The sum of natural numbers up to n and the sum of cubes natural numbers up to n. So what I mean by that is just if we could 1, 2, 3, and so on all the way to some last number we will call "n". And we added them all, what is a formula for that? In other words, what is 1 + 2 + 3 + ... + (n - 2) + (n - 1) + n in terms of n? The formula for this is (n + 1) * (n / 2). If you want a proof for why this formula works, just tell me.
Now we also need to know the sum of cubes formula which is also adding the same counting numbers up to n but we are putting each one to the 3rd power as well. So 1^3 + 2^3 + 3^3 + ... + (n - 2)^3 + (n - 1)^3 + n^3. The formula for this is [ (n + 1) * (n / 2) ]^2 which again, if you want to know why then ask me but as you can see, this formula so happens to be the formula for the last part but squaring the entire thing as well.
The problem shows the sum of cubes inside of a square root, so plugging in our sum of cubes formula, the square abs square root cancel and we are left with the original sum of natural numbers formula (n + 1) * (n / 2). So it really just goes to 1 + 2 + 3 + ... + 2023 + 2024.
A cube is a sum of "2-D honeycomb" numbers, so each difference is each "honeycomb" number:
1 + 6*triangle(x-1)
Sum of consecutive honeycomb numbers, or a cube:
x + 6*triangle^2(x-1)
Sum of consecutive cubes:
triangle(x) + 6*triangle^3(x-1)
or
x(x+1)/2 + x(x-1)(x+1)(x+2)/4
(2x(x+1) + (x^3-x)(x+2))/4
(2x^2+2x + x^4+2x^3-x^2-2x)/4
(x^4+2x^3+x^2)
x^2(x^2+2x+1)/4
(x(x+1)/2)^2
Sum k^3 = (n(n+1)/2)^2, that is why
Knew it was 2049300 from the start due to the sum of cubes being the square of the corresponding triangle number, so it was just the 2024th triangle number
stair number is more accurate name tho
@@daanwinne2596 I disagree
Assuming is nice, but no proof (even though that in this case that's true). I can create a series that will start like that, 1, 2, 3, 4, 5 and the next number will be 126 not 6. If someone wonders what it could be, here's an example:
(X-1)(X-2)(X-3)(X-4)(X-5)+X
But I didn't say that I would prove it ¯\_(ツ)_/¯
The goal was to demonstrate the pattern
Just do the 1+2+…100 thing
it will equal 2025*1012 which is around 2 million
Thing😂, sorry but the why you said it it's funny.
answer: 2,049,300
I don’t know, probably))
2025 x 1012
Yes
@@martonbalazskajari5424 Gauss Summation rulez! :)
This can be proved using mathematical induction and the fact that the sum of 1+2+3+...n = n(n+1)/2
Assuming that the equation holds for n numbers. Then 1³ + ... + n³ + (n+1)³ = (n*(n+1)/2)² + (n+1)³. = (n+1)²(n+2)²/4. Take the square root and both sides, and voila then you have proved that the inductive step follows. Thus, the equation must hold for all natural numbers.
This is shown most easily by geometric argument. The numbers can be represented as squares on a grid. Those squares can always be combined whole (odd terms) or cut in half (even terms) to form one large square. Each time a new term is added the large square's side length increases accordingly:
one 1x1 square
plus two 2x2 squares
plus three 3x3 squares
plus four 4x4 squares
etc...
Just because it works for the first few numbers doesn't mean it works for all numbers, so it's not a solution unless you can prove it with a general formula
But I never said that it was a solution) the point was to give the idea
Sum of cube of n natural numbers = ((n(n+1))/2)² . In this case the square root and perfect square cancel, so we get (n(n+1))/2, which is the sum of n natural numbers.
Ok just ignore the 1
2^3=2x2x2=8
So 2x2=4 is a perfecr square and then doubling again 4x2=8 then adding previous number 1+8=9.
So it's just coincidence it's more about psychological
It's not a coincidence. The derivative of x³ is 3x² so the difference between every cubed whole number is always a multiple of a square whole number, and the whole sequence being in a square root function turns all those x² terms into x
Lots of ways to do it. There are inductive methods, but then there's also the calculus of finite differences.
Make a generating function of the series, put a bound of 2024 terms, take a square root
This is very much sub imo level why tf are you using generating functions
Knowing the pattern, it gets easily solved by sum of arithmetic progression series and answer comes 2049300
These lessons should be sung by Anselmo during Pantera songs. 21 is the next number.
Well i know the formula for the sum of cubes from my further maths a level, i also know it happens to be the sum of the natural numbers up to the same limit all squared as that makes it easier to remember.
So I'd simplfy it to 1 + 2 + 3 + ... + 2024 which is (1/2)(2024)(2025)
I am a learner but I will try to share my knowledge to human society.
素晴らしい解説
Thank you!
Maybe we could prove this by induction
It screams induction
@@serk-s I managed to do it! Induction's about as far as my proof skills go
@@capsandnumbers In such questions induction is the way to go. So congrats :D
Solve this question
Limit x approaches to 0 e^ax - e^bx / 2x
knowing to prove it is more important than just knowing the trick, or else you wouldn't see the beauty behind!
We can prove it by using discrete integration
go ahead
@@Insigniume
the default here is Σn mean summation of x from x=1 to x=n . Just for simpler illustration.
Δ(Σn)²
=[Σ(n+1)]²-[Σn]²
=(Σ(n+1)-Σ(n))(Σ(n+1)+Σ(n))
=(n+1)(2Σ(n)+n+1)
=(n+1)(n(n+1)+(n+1)) {useΣn=0.5n(n+1)}
=(n+1)(n+1)(n+1)
=(n+1)³
now forget the above default
so ΣΔ[Σ(n)]²=Σ(n+1)³
(Σ(n+1))²-(Σ(0))²=Σ(n+1)³
(Σn)²=Σn³
i used finite differentiation
@@hiyayahiyaya5645 that's what I was talking about, didn't know it was called finite in English, we call it discrete in French
@@vonwthaud289 i can understand what you are talking about , i haven't challenged you .
Because if you cube them (power 3) and then square root them (power 2) it gives you an exponent of 1 since 3-2=1
But it only works when you separately take the square roots of the cubes of the numbers. Here, the square root is taken over the whole sum of the cubes in which case it doesn't work. Though when its exponential homogeneity is concerned it works.
But it only works when you separately take the square roots of the cubes of the numbers. Here, the square root is taken over the whole sum of the cubes in which case it doesn't work. Though when its exponential homogeneity is concerned it works.
Actually we can write 1³+2³+........ = (1+2.....)² therefore you just cancelled out the square root with the square.
The problem: "Oh? Youre approaching me?"
Why is there not a symbol for 1+2+..+n, like we have factorial, surely there could be some sumorial symbol. 2048# ?
The general case is solved by faulhabers formula (which to me feels so weird to be true but well, it is).
Here is a short list: en.m.wikipedia.org/wiki/Faulhaber%27s_formula#Examples
The summation formula that my calculus 1 professor gave me on our integration test was true!!!!
Did he just give you the formula or did he prove it?
@@THEMATHSTER1 he just gave it to us for calculating Riemann sums. The next class did summations and he knew we’d experience the misery then.
Just use the formula
[n(n+1)/2]² and simplify it's prime factors to square root
Prove through induction that the sum of the cubes of the first n numbers is ( (n+1)*n / 2 )^2 and that's basically it
Thats the beauty of mathematics
Does no one know sumission n³ here
Yes, you can get the formula from induction. Slight correction, it's summation
I don't understand
@@noname-ed2un he's referring to the sum 1^3+2^3+3^3+...+n^3, which is equal to (n(n+1)/2)^2
1^3+2^3+________+n^3=
=(n×(n+1)/2)^2 and we know that n=2024 then we get square root of (2024×2025/2)^2 after after destroying root and square we get 2024×2025/2 and it equals to the 1012×2025 easy
This works bc sigma of x³=x²(x+1)²÷4 and taking the square root would give us x(x+1)÷2 wjich is equal to sigma of x
And what if the pattern stops holding up after some number of cases? Take Borwein integral for example. You didn't actually calculate the number your problem requires since you haven't provided any proof of the pattern, like induction.
Круто, что вы и на английском делаете ролики!
(1+2+3+...+n)^2 =
(n(n+1)/2)^2 =
(n^2)(n+1)^2/4 =
(n^2)(n^2+2n+1)/4 =
¼n^4+½n^3+¼n^2
Answer is 20,49,300
We know that , 1^3 +2^3 +3^3 +.......+2024^3 =(2024)^2 ×(2025)^2 /2^2
Since √2024×2024×2025×2025/4
=2024×2025/2
=20,49,300
it removes the exponent 2 in the exponent 3
if you get what i mean
int add = 0;
int number = 0;
do {
number++;
add = add (number*number);
} while (number
Wtf?
@@susmitadutta1029 java
Bruh C++?
This won't work when int overflows.
Why Java :(
I got 8,291,469,824 idk if that’s correct or not.
Nope
@@N4chiket recalculated is it 282,978,244.48
@@paytonrickle6785so... 1+2+3+...+n has a formula in it. The formula is n(n+1)/2. If the lowest number is not one, just do (number of terms)(smallest term+largest term)/2 for continuous numbers In this case, it's 2024(1+2025)/2. which leads up to 2049300.
@@paytonrickle6785 wrong again, the answer is 2,049,300. The formula for the sum of the first n natural numbers (1, 2, 3, etc) is n(n+1)/2. I have no clue what you are doing to get your answer.
{3+8}={11 +9} {=20 +12}={32 +15}={47+ 18}=63 8144= 15714.2 10^10 3^19 2^7 .2^1 2^52^5 3^19^1 1^1.1^1 1^12^13^1^1 23 (x ➖ 3x+2)
a1=1, a2=2, a3=3.........a2024=2024
The sum of all is S
S=(a1+a2024)×2024÷2=2025×1012
The question at the end can be easily proven by induction. But I did'nt dare to prove it directly.
Beautiful sense
Use summation r^3
Math is amazing!
Prove it for small number of terms then prove that if it’s true for a given number of terms it’s also true for that many terms plus one more
Prove by induction, n=1 is trivial
For step n->n+1 say \sigma k^3 = (\sigma k)^2
Notice now how when adding one term we get
\sigma k^3 + (n+1)^3 =?= (\sigma k + n+1)^2
Note how when expanding the right side we get the term from the induction hypothesis, and two terms relating to n+1
Canceling using hypothesis we get
(N+1)^3 =?= n+1((2\sigma k)+n+1)
Or equiv
((n+1)^2 - (n+1)) /2 =?= \sigma k
Prove this again by induction, where n= 0 is trivial
Etc etc now I would need to substitute bounds RUclips comments have no markup
Disgusting mess = sigma k^3
You could do a proof by induction
2,049,300 is the answer for anyone wondering
I didn’t even think of the sum of the cuberoot(term), I just thought “quadratic sequence” 😭
It's called induction.
Induction
But still don't know how I would discover it
Not necessarily )
It could be proven using formulas of the sum of n natural numbers and the sum of their cubes.
The question is how do you prove these formulas))
0@@THEMATHSTER1its easy to prove , take a polynomial (x+1)³ expand it for each term till 1³( 1³ , 2³ , ......x³) and now replace x with x-1 everywhere and when you add all the equations you are left with x³ = some equation because rest everything gets cancelled and then you will also have some term (1²+2²+.......x²) use the formula for it also , take n common do some solving and youre done
Yes, to prove those... Induction
1) show that it works for 1
2) prove that if it works for k then it works for k+1
Not really difficult, but I'm saying I would need to have an observation or an epiphany about the formula beforehand in order to use induction
And when n=infinity the answer is ofc *-1/12*
Ans is 2049300, average Grade 6 question in India
Why does this happen? Amazing
The question is why does this work 😍😍
2049300
So what is the answer for 1+2+3 +...+ 2024 ?
1012*2025 that's 2049300. That's how I would had approached it
a³+b³=( a+b)²
This means you can solve it by induction!
The method is interesting, but the coincidence doesn't prove it will work for every element under the root sign, so mathematically it is not that correct
Did I say that I wanted to prove something? :)
The goal was to demonstrate the idea and the method
Sum of cubes of first n natural numbers is equal to the square of sum of first n natural numbers
2 049 300
If that always works, the right answer is 2,049,300
its just n(n+1)/2 im lazy to prove it 😅
Fym, showing how to get from 1+2+3+…+n=n(n+1)/2 is very easy
S=1 +2 +3+… +n
S=n+(n-1)+(n-2)+…1
S+S=(n+1)+(n+1)+(n+1)+…+(n+1)
^n-times
2S=n(n+1)
S=n(n+1)/2
But the video isn’t even about that, it’s about using intuition to solve what would be a very tedious problem without it
@@SilentGamer._ no no the some of natural numbers cubes of 1 to n is actually {(n+1)n/2}^2
2,049,300
Ive seen some number sequence that break pattern in 5th term. So its not a reliable way.
It's not a reliable way of what?
Of proving things? Yes. But I didn't prove anything here. The goal was to DEMONSTRATE
2024x2025/2=2,049,300
Yes that's so sigma
Ճ=1012×2025==2.025.000+20.250+4.050=
=2.049.300 !!!
М=(2024÷2)×(1+2024) !!!😊❤😮😮😮😮😮❤
Ans is 2,049,300
The sum of 1³ to n³ is always a square? What? Why?
Something about induction.
Well, you can PROVE it with induction, if that's what you mean.
@@Debaser36 I know.
Simply proving the formula in the first place is very difficult, unlike the formula for triangle numbers.
It's possible to visualise it however. A 1 x 1 square is placed at the top left corner, then we take 2^3 and split it into 2 x 2^2, in the form of 2 separate 2x2 blocks.
One block is simply placed diagonally to the 1 x 1 block, something like this:
1XX
X22
X22
Then, the other 2 x 2 block is split into 2 separate 1 x 2 blocks, which are then placed in the gaps above and to the left of the 2 x 2 block, creating a full 3 x 3 block, proving that 1^3 + 2^3 = (1+2)^2 = 3^2 = 9
122
222
222
Then 3 3x3 blocks are simply placed in an L-shape around the current set of blocks like so:
122333
222333
222333
333333
333333
333333
This shows that 1^3 + 2^3 + 3^3 = (1+2+3)^2 = 6^2 = 36
Then 4 4x4 blocks are then used, where 3 of the 4 x 4 blocks are placed in an L-shape like so:
122333XXXX
222333XXXX
2223334444
3333334444
3333334444
3333334444
XX44444444
XX44444444
XX44444444
XX44444444
The remaining 4 x 4 block is split into 2 separate 2 x 4 blocks which then fill in the gaps, creating a block of (1 + 2 + 3 + 4) x (1 + 2 + 3 + 4), and so on.
Thus visually proving that the sum of the first n cubes is the square of the sum of the first n numbers (i.e. the square of the nth triangle number).
However, trying to represent this algebraically is difficult.
It has something to do with splitting cubes into a product of squares and the original number, such that n^3 = n(n^2), but every even number cube is split further, such that (2n)^3 = (2n)(2n)^2 = (n/2 + n/2)(2n)^2
i.e.:
2^3 = 2 x 2^2 = (1+1) x 2^2
4^3 = 4 x 4^2 = (2+2) x 4^2
6^3 = 6 x 6^2 = (3+3) x 6^2
8^3 = 8 x 8^2 = (4+4) + 8^2
So, it's something like:
1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 =
1x(1^2) + (1+1)x(2^2) + 3x(3^2) + (2+2)x(4^2) + 5x(5^2) + (3+3)x(6^2) and so on... but whether the last term os n x n^3 or (n/2 + n/2) x (n^3) depends on whether n is even or odd in the visual proof.
because sum of 1 to n cubes is equal to square of sum of 1 to n
"But why does this happen?"
Uhhhh uhhhhh PI?
Some kinda progression
Day 69 preparing for ssc cgl
But you can use AP
i knew how to do it before u explained
2.049.300
I would run away
why does it work?
Fairly trivial to prove, just do induction lmao
1^3+2^3+3^3.....=(1+2+3...)^2
Proof via full Induktion ...
Bro this a basic in our 10th grade in india😅
You mean this specific problem? Why? How often do you need to square root a sum of consecutive cubes?
((2024×2025)/2)²
2 049 300? is it right?
yes
sqrt(2024!)
2073600