Mathematics Thread

[Ranzha]

So today I was at this school function thingmajig.

So I just went in and 1/0

Needless to say the school isn't there anymore. I was barely able to escape; good thing I am a dragon.

I thought you were a...
Cool Frog.

Spoiler

NOOOOOOOOOOOO

 

[asportking]

oh. that must be what I'm doing right now. gahhhh... i am hating on whoever invented proofs. I mean, they're a great way to organize your thoughts and stuff, but require theorems for reasoning, and just takes too long. I'm in 8th grade, and I heard the geometry in high school is way easier, with barely any proofs. Not fair.

Yeah, I didn't really like proofs either, but it's a lot more useful in real life. Like, in class, you're told to prove that two angles are the same. In the real world, you don't even know if the two angles are the same, and just saying "they look about the same" isn't going to cut it. They make you show proofs so that you can understand your reasoning behind things.

 

[ben1996123]

How do you prove this?

\( \sum_{n=1}^{x}n^{2}=\frac{x(x+1)(2x+1)}{6} \)

 

[tozies24]

How do you prove this?

\( \sum_{n=1}^{x}n^{2}=\frac{x(x+1)(2x+1)}{6} \)

mathematical induction on n.

here is an outline of what induction is:
There are two steps.

1. Prove that the base case works.
2. Assume that is works for the nth term. so what you have above. Then you want to show that it works for n+1.

So for your problem, 1^2 = 1= 1(1+1)(2*1+1)/6 = 1. So the base case works.

Now we take the inductive step: assume that it works for n. We want to show it works for n+1.
So we have SUM : (n+1)^2 = n^2 +2n + 1. BUT we know what n^2 is by the induction hypothesis.
So (n+1)^2 = n^2 + 2n + 1 = n(n+1)(2n+1)/6 + 2n+1. Then you can do some algebra and you will find out that that equals (n+1)(n+1+1)(2(n+1)+1)/6. So you have showed that this is true for all n. END PROOF.

Hopefully this helps, sorry for not using fancy math computer writing too.

 

[Cool Frog]

I thought you were a...
Cool Frog.

Spoiler

NOOOOOOOOOOOO

I have legs of a frog, Body of a catapilla, and soul and mind of a dragon.

 

[ben1996123]

mathematical induction on n.

here is an outline of what induction is:
There are two steps.

1. Prove that the base case works.
2. Assume that is works for the nth term. so what you have above. Then you want to show that it works for n+1.

So for your problem, 1^2 = 1= 1(1+1)(2*1+1)/6 = 1. So the base case works.

Now we take the inductive step: assume that it works for n. We want to show it works for n+1.
So we have SUM : (n+1)^2 = n^2 +2n + 1. BUT we know what n^2 is by the induction hypothesis.
So (n+1)^2 = n^2 + 2n + 1 = n(n+1)(2n+1)/6 + 2n+1. Then you can do some algebra and you will find out that that equals (n+1)(n+1+1)(2(n+1)+1)/6. So you have showed that this is true for all n. END PROOF.

Hopefully this helps, sorry for not using fancy math computer writing too.

How would you find the equation if you didn't know that you were specifically looking for \( \frac{n(n+1)(2n+1)}{6} \)?

 

[tozies24]

How would you find the equation if you didn't know that you were specifically looking for \( \frac{n(n+1)(2n+1)}{6} \)?

Probably just look at it and look for a pattern? I am not sure what you are asking.
I guess you would look at what each of the terms of the sequence of partial sums is. So look at the sequence of the finite sum of n^2.

1, 5, 14, 30, 55, 91, etc.

Then you could find a function that describes these outputs where the function goes from the natural numbers to the natural numbers.
Even though this would be cheating, you could plug these values into a graphing calculator and hit the button for cubic line or whatever its called and then you would get the equation (2n^3+3n^2+n)/6.

I am not sure how the people who found these equations did it (either guess and check or something I don't know). Maybe they just looked at it for a while and then wrote down the answer because they were that brilliant.

 

[qqwref]

cubingawsumness/asportking: The "proofs" they have you do in geometry are definitely more work than they need to be, but don't let that put you off the idea of proofs themselves. In actual math you don't have to explain things in extreme detail and justify every single step with a theorem, so proofs are often short and elegant, explaining why something is true when it's not immediately clear. In fact, all of the higher math (that is, the more interesting stuff) is based off proving things to be true.

ben1996123: If you're wondering about the motivation for that expression, one way to try to find it would be to write out a sequence of sums-of-squares, and then take finite differences (see wikipeda) a few times to realize that the sequence seems to be a cubic polynomial. (Finite differences would let you see if any expression works out to a polynomial of integers.) From there you could either do a system of equations or work through induction to get the actual equation. In general, problems of sequences and differential equations are a lot easier when you can guess the general form of the answer.

 

[samehsameh]

ben1996123: I wrote out a solution for you then i saw that qqwref posted how to do it but so i didnt waste my time here it is anyway.

Spoiler

 

[ben1996123]

ben1996123: I wrote out a solution for you then i saw that qqwref posted how to do it but so i didnt waste my time here it is anyway.

Spoiler

Thanks, yay matrices. I guess I'll learn about matrices tomorrow so I can understand it

 

[cubernya]

Has anyone ever proved using maths that god'd number is 20?

 

[jeff081692]

Has anyone ever proved using maths that god'd number is 20?

I found the history interesting.
http://www.cube20.org/

 

[samehsameh]

ben1996123: For the purpose of Gaussian elimination (which is what i did) they're just tables of coefficients. see the 4 equations above the first matrix, thats the set of equations im trying to solve, i had 4 unknowns in my cubic equation so i needed 4 equations to solve them. So column 1 represnts contant a, column 2 represents b, column 3 represents c and column 4 represents d. Im sure you could solve the equations 2x+3y=13 and x-2y=-4 simultaneously. You eliminate one variable then solve for the other then substitute that back into the original to find the variable you eliminated. This is essentially Gaussian Elimination on 2 equations in 2 unknowns. If you zoom into the picture, while it is blury, they're little notes like r4 - 64 x r1 which are just what im doing, i.e. im taking row 1 away from row 4 64 times, eliminating the first variable. The vertical line to the right of the matrices are basically = signs. So you read the first row of the matrix as a + b + c + d = 1 etc. In the 3rd matrix, i make it so that the first non zero element of each row is equal to 1, this makes it easier to eliminate them in the next step since i can do r3-r2 instead of r3 -(4.5) x r2. Remember what i do to the first element i do the the others on that row, with the corresponding element in the other row. The idea of Gaussian Elimination is to end up with the identity matrix (a square matrix with 1's on the diagonal, zeros every where else) excluding the numbers after the vertical line. You do this by eliminating down the equations which is what i did, then elimating up the equations, i stopped before eliminating upwards and used backwards substitution instead because its only 4 equations, huge matrices i would suggest finishing Gaussian elimination. Hope that helps getting you started in matrices.

 

[Christopher Mowla]

@samehsameh and qqwref, I don't like that approach.


Ben, I found this problem interesting because I always just saw it proved with induction.

With some effort and luck, I made my own direct proof.

It took me several hours to complete it, but the math behind it is very simple. Here is another direct proof that I found after I made mine, but I think mine is much easier to understand.

We need to show that \( \sum\limits_{i=1}^{n}{i^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \)

Proof (Direct)

Spoiler

Observe the following graph.

Notice that the areas of the squares which have vertices from the point (0,0) to the points (1,1), (2,2), (3,3), and (4,4) are (1)(1) = 1, (2)(2) = 4, (3)(3) = 9, and (4)(4) = 16, respectively.

Representing the area of each square in terms of the area of the square before it,
\( \left( 1 \right)\left( 1 \right)=\left( 1 \right)\left( 1 \right) \)
\( \left( 2 \right)\left( 2 \right)=\left( 1 \right)\left( 1 \right)+3\left( 1 \right)\left( 1 \right) \)
\( \left( 3 \right)\left( 3 \right)=\left( 2 \right)\left( 2 \right)+5\left( 1 \right)\left( 1 \right)=\left( 1 \right)\left( 1 \right)+3\left( 1 \right)\left( 1 \right)+5\left( 1 \right)\left( 1 \right) \)
\( \left( 4 \right)\left( 4 \right)=\left( 3 \right)\left( 3 \right)+7\left( 1 \right)\left( 1 \right)=\left( 1 \right)\left( 1 \right)+3\left( 1 \right)\left( 1 \right)+5\left( 1 \right)\left( 1 \right)+7\left( 1 \right)\left( 1 \right) \)
\( \vdots \)
\( \left( n \right)\left( n \right)=... \)
\( \text{So }\sum\limits_{i=1}^{n}{i^{2}}=\left( 1 \right)\left( 1 \right)+\left[ \left( 1 \right)\left( 1 \right)+3\left( 1 \right)\left( 1 \right) \right]+\left[ \left( 1 \right)\left( 1 \right)+3\left( 1 \right)\left( 1 \right)+5\left( 1 \right)\left( 1 \right) \right]+\cdot \cdot \cdot \)
\( =1+\left( 1+3 \right)+\left( 1+3+5 \right)+\left( 1+3+5+7 \right)+\cdot \cdot \cdot \)

We can see from the equation above that there will be n 1's, \( \left( n-1 \right) \) 3's, \( \left( n-2 \right) \) 5's, etc., in the summation.

\( =1\left( n-0 \right)+3\left( n-1 \right)+5\left( n-2 \right)+7\left( n-3 \right)+9\left( n-4 \right)+\cdot \cdot \cdot \)

\( =\sum\limits_{i=1}^{n}{\left( 2i-1 \right)\left( n-\left( i-1 \right) \right)} \) \( =\sum\limits_{i=1}^{n}{\left( 2i-1 \right)\left( n-i+1 \right)} \) \( =\sum\limits_{i=1}^{n}{-2i^{2}+n2i+3i-n-1} \) \( =\sum\limits_{i=1}^{n}{-2i^{2}+\left( 2n+3 \right)i-\left( n+1 \right)} \)
\( =-2\sum\limits_{i=1}^{n}{\left[ i^{2} \right]}+\left( 2n+3 \right)\sum\limits_{i=1}^{n}{\left[ i \right]}-\left( n+1 \right)\sum\limits_{i=1}^{n}{\left[ 1 \right]} \)
\( =-2\sum\limits_{i=1}^{n}{\left[ i^{2} \right]}+\left( 2n+3 \right)\left[ \frac{n\left( n+1 \right)}{2} \right]-\left( n+1 \right)\left[ n \right] \)
\( =-2\sum\limits_{i=1}^{n}{\left[ i^{2} \right]}+\frac{n\left( n+1 \right)\left( 2n+3 \right)-2n\left( n+1 \right)}{2} \)
\( =-2\sum\limits_{i=1}^{n}{\left[ i^{2} \right]}+\frac{n\left( n+1 \right)\left[ \left( 2n+3 \right)-2 \right]}{2} \)
\( \sum\limits_{i=1}^{n}{i^{2}}=-2\sum\limits_{i=1}^{n}{\left[ i^{2} \right]}+\frac{n\left( n+1 \right)\left( 2n+1 \right)}{2} \)
\( \Rightarrow \left( 1+2 \right)\sum\limits_{i=1}^{n}{i^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{2} \)
\( \Rightarrow 3\sum\limits_{i=1}^{n}{i^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{2} \)
\( \Rightarrow \sum\limits_{i=1}^{n}{i^{2}}=\frac{1}{3}\frac{n\left( n+1 \right)\left( 2n+1 \right)}{2} \)
\( \Rightarrow \sum\limits_{i=1}^{n}{i^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \)

Therefore, \( \sum\limits_{i=1}^{n}{i^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \)
Q.E.D.

 

[qqwref]

Shortening of cmowla's proof:

Spoiler

Proof
Using cmowla's image, we can see that
\( n^2 = \sum\limits_{i=1}^{n}{2i-1} \)

Let
\( k = \sum\limits_{j=1}^{n}{j^2} \)
Then we can write
\( k = \sum\limits_{j=1}^{n} \sum\limits_{i=1}^{j} {2i-1} \)
\( = \sum\limits_{i=1}^{n} \sum\limits_{j=i}^{n} {2i-1} \)
\( = \sum\limits_{i=1}^{n} {(n-i+1)(2i-1)} \)
\( = \sum\limits_{i=1}^{n} {\left(-2i^2 + (2n+3)i - (n+1)\right)} \)
\( = -2 k + (2n+3) \sum\limits_{i=1}^{n}\left(i\right) - (n+1)\sum\limits_{i=1}^{n}\left(1\right) \)
\( = -2 k + (2n+3) \frac{n(n+1)}{2} - n(n+1) \)
\( = -2 k + n(n+1)(\frac{2n+3}{2} - 1) \)
\( = -2 k + \frac{n(n+1)(2n+1)}{2} \)

Assuming k exists and is finite (very reasonable assumptions), we can directly solve for it:
\( k = -2 k + \frac{n(n+1)(2n+1)}{2} \)
\( 3k = \frac{n(n+1)(2n+1)}{2} \)
\( \sum\limits_{j=1}^{n}{j^2} = k = \frac{n(n+1)(2n+1)}{6} \)

 

[Christopher Mowla]

Haha, have fun qqwref!

I have proved the formula for \( \sum\limits_{i=1}^{n}{i^{3}} \) in a similar manner. I then went to see about \( \sum\limits_{i=1}^{n}{i^{4}} \) and I found the formula for that one too.

In fact, I found a pattern.

This probably existed for several hundred years already, but here is a recursion formula I ultimately ended up with.

\( \sum\limits_{i=1}^{n}{i^{R}}=\sum\limits_{i=1}^{n}{\frac{\left( i-n-1 \right)\left( i-1 \right)^{R}+\left( R-i+n+1 \right)i^{R}}{R+1}} \)
​

, where R is the power of the sum. It works for R = 1 to infinity (it doesn't work for \( \sum\limits_{i=1}^{n}{i^{0}} \)).


Note that, even though the recursion formula has R (when it should be R-1), after you substitute a number in for R and expand, the highest exponent remaining will always be R-1.

I made it using the "trick" of my proof along with seeing how to derive the higher power sums (greater than 2).

For example, suppose we want to find the formula for \( \sum\limits_{i=1}^{n}{i^{4}} \).

Substitute 4 for R in the series.

\( \sum\limits_{i=1}^{n}{i^{4}}=\sum\limits_{i=1}^{n}{\frac{\left( i-n-1 \right)\left( i-1 \right)^{4}+\left( 4-i+n+1 \right)i^{4}}{4+1}} \)

\( \sum\limits_{i=1}^{n}{i^{4}}=\frac{1}{5}\sum\limits_{i=1}^{n}{\left[ i^{3}\left( 4n+10 \right)-i^{2}\left( 6n+10 \right)+i\left( 4n+5 \right)-\left( n+1 \right) \right]} \) (see, \( i^{3} \) is the highest degree term).

\( =\frac{1}{5}\left( \left[ \frac{n^{2}\left( n+1 \right)^{2}}{4} \right]\left( 4n+10 \right)-\left[ \frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right]\left( 6n+10 \right) \right) \) \( +\frac{1}{5}\left( \left[ \frac{n\left( n+1 \right)}{2} \right]\left( 4n+5 \right)-\left( n+1 \right)\left[ n \right] \right) \)

\( =...=\frac{n\left( n+1 \right)\left( 2n+1 \right)\left( 3n^{2}+3n-1 \right)}{30} \)

 

[aronpm]

Let's see how I do with [noparse]\( [/noparse]

First let's find \( \sum\limits_{i=1}^{n}{i} \)
\( \sum\limits_{i=1}^{n}{i} = 1 + 2 + ... + (n-1) + n = n + (n-1) + ... + 2 + 1 \)

\( 2\sum\limits_{i=1}^{n}{i} = (n+1) + (n-1+2) + ... + (2+n-1) + (1+n) \)

This (n+1) term is repeated n times, so
\( 2\sum\limits_{i=1}^{n}{i} = n(n+1) \)

\( \sum\limits_{i=1}^{n}{i} = \frac{n(n+1)}{2} \)

Also, it's fairly trivial that \( \sum\limits_{i=1}^{n}{1} = 1+1+...+1+1 = n \)

To find the sum of squares we'll consider the sum of cubes:
\( \sum\limits_{i=1}^{n}{i^3} = 1^3 + 2^3 + ... (n-1)^3 + n^3 \)

\( \sum\limits_{i=1}^{n}{(i-1)^3} = 0^3 + 1^3 + ... (n-2)^3 + (n-1)^3 \)

Taking the difference of the two yields:
\( \sum\limits_{i=1}^{n}{i^3} - \sum\limits_{i=1}^{n}{(i-1)^3} = 1^3 + 2^3 + ... n^3 - (1^3 + 2^3 + ... + (n-1)^3) = n^3 \)

If we re-arrange, expand and combine:
\( n^3 = \sum\limits_{i=1}^{n}{(i^3 - (i^3 - 3i^2 + 3i - 1))} = \sum\limits_{i=1}^{n}{(3i^2 - 3i + 1)} \)

Expanding the sum we get
\( n^3 = 3\sum\limits_{i=1}^{n}{i^2} - 3\sum\limits_{i=1}^{n}{i} + \sum\limits_{i=1}^{n}{1} = 3\sum\limits_{i=1}^{n}{i^2} - \frac{3n(n+1)}{2} + n \)

Now we can solve for the sum
\( 3\sum\limits_{i=1}^{n}{i^2} = n^3 + \frac{3n(n+1)}{2} - n = n^3 + \frac{3}{2}n^2 + \frac{1}{2}n \)

\( \sum\limits_{i=1}^{n}{i^2} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} = \frac{n(2n^2 + 3n + 1)}{6} = \frac{n(2n+1)(n+1)}{6} \)


I think this is easier to understand than the other proofs given

edit: btw it's not my proof. It was the one the lecturer showed us in my calculus lecture today. \)

 

[Christopher Mowla]

\( 2\sum\limits_{i=1}^{n}{i} = (n+1) + (n-1+2) + ... + (2+n-1) + (1+n) \)

How did your instructor say he got that \( 2\sum\limits_{i=1}^{n}{i}=n\left( n \right)+n\left( 1 \right)=n\left( n+1 \right) \)? Yes they turn out to be equal, but I cannot see any justification. This right hand side of that equation is showing the result you are trying to prove.

Taking the difference of the two yields:
\( \sum\limits_{i=1}^{n}{i^3} - \sum\limits_{i=1}^{n}{(i-1)^3} = 1^3 + 2^3 + ... n^3 - (1^3 + 2^3 + ... + (n-1)^3) = n^3 \)

Again, it is not immediately obvious why \( \sum\limits_{i=1}^{n}{i^{3}}-\sum\limits_{i=1}^{n}{\left( i-1 \right)^{3}}=n^{3} \) (yes they are equal, but how is that shown?). I mean, he did truncate off the zero from \( \sum\limits_{i=1}^{n}{\left( n-1 \right)^{3}} \), but then when you match the terms to subtract, you must add one more term to be able to subtract the first n cubes from each other.
\( 1^{3}+2^{3}+\cdot \cdot \cdot +n^{3}-\left( 1^{3}+2^{3}+\cdot \cdot \cdot +\left( n-1 \right)^{3}+n^{3} \right)=0 \)
.
Even if I didn't add the \( n^{3} \) back,

\( 1^{3}+2^{3}+\cdot \cdot \cdot +n^{3}-\left( 1^{3}+2^{3}+\cdot \cdot \cdot \left( n-1 \right)^{3} \right)=n^{3}-\left( n-1 \right)^{3}\ne n^{3} \).

I cannot see how you can know that \( \sum\limits_{i=1}^{n}{i^{3}}-\sum\limits_{i=1}^{n}{\left( i-1 \right)^{3}}=n^{3} \).

The way I see it, your instructor showed how you prove that \( \sum\limits_{i=1}^{n}{i^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \) already knowing (not proving at all) that \( \sum\limits_{i=1}^{n}{i^{3}}-\sum\limits_{i=1}^{n}{\left( i-1 \right)^{3}}=n^{3} \).

 

[Christopher Mowla]

Here's a direct proof I just came up with for \( \sum\limits_{i=1}^{n}{i}=\frac{n\left( n+1 \right)}{2} \).

We need to prove that \( \sum\limits_{i=1}^{n}{i}=\frac{n\left( n+1 \right)}{2} \).
Proof

Spoiler

Consider adding an odd number of consecutive integers.

\( 1+2+3=\left( 3 \right)2 \)
\( 1+2+3+4+5=\left( 5 \right)3 \)
\( 1+2+3+4+5+6+7=\left( 7 \right)4 \)
...

(Where the number in parenthesis is the last/largest integer).

Let n represent the odd number of consecutive integers which we add.

Let's write ordered pairs based off of the multiples of the largest number (which is an odd number).

Note that the ordered pairs are of the form:
(Number of consecutive integers, multiple of the largest integer added)

\( \left\{ \left( 3,2 \right),\left( 5,3 \right),\left( 7,4 \right) \right\} \)

\( \frac{3-2}{5-3}=\frac{4-3}{7-5}=\frac{1}{2}=m \)

Just using the point \( \left( 7,4 \right) \) and the slope, we can write a linear equation which satisfies both.

\( y-4=\frac{1}{2}\left( n-7 \right)\Rightarrow y=\frac{1}{2}n-\frac{7}{2}+4=\frac{1}{2}n-\frac{7}{2}+\frac{8}{2}=\frac{1}{2}n+\frac{1}{2} \).

\( =\frac{1}{2}\left( n+1 \right) \)

(Note that we can be certain that the slope is constant because the integers we are adding increase by one each time).

So \( f\left( n \right)=\frac{1}{2}\left( n+1 \right) \) gives us the multiples of the largest (last) integer added (which is odd), not the sum.

Recall that n represents the last/largest integer in the sum (see the original sums of integers). So the sum of the first n integers (where n is an odd number) is the following (refer back to the beginning):

\( n\left( \frac{1}{2}\left( n+1 \right) \right)=\frac{n\left( n+1 \right)}{2}. \)

-------
If we instead add an even number of consecutive integers (that is, n is an even number).

\( 1+2=\left( 1 \right)+1 \)
\( 1+2+3+4=\left( 1+2+3 \right)+4 \)
\( 1+2+3+4+5+6=\left( 1+2+3+4+5 \right)+6 \)

Clearly the sum will be (based on our findings for an odd number of consecutive integers):

\( \frac{\left( n-1 \right)\left( \left( n-1 \right)+1 \right)}{2}+n=\frac{n\left( n+1 \right)}{2} \).

Therefore the sum of both an odd number and even number of consecutive integers is \( \frac{n\left( n+1 \right)}{2} \).

Therefore \( \sum\limits_{i=1}^{n}{i}=\frac{n\left( n+1 \right)}{2} \).

Q.E.D.

 

[aronpm]

How did your instructor say he got that \( 2\sum\limits_{i=1}^{n}{i}=n\left( n \right)+n\left( 1 \right)=n\left( n+1 \right) \)? Yes they turn out to be equal, but I cannot see any justification. This right hand side of that equation is showing the result you are trying to prove.

Uh... isn't it... kind of... obvious? For each term (n-i) you've got a term (i+1) so you add them together to get (n+1), and each sum had n terms. I didn't see any justification when you said

We can see from the equation above that there will be n 1's, (n-1) 3's, (n-2) 5's, etc., in the summation.

Again, it is not immediately obvious why \( \sum\limits_{i=1}^{n}{i^{3}}-\sum\limits_{i=1}^{n}{\left( i-1 \right)^{3}}=n^{3} \) (yes they are equal, but how is that shown?).

Again, how is that not obvious? You've got the first n cubes, and you take the first n-1 cubes away. What do you have left?

Spoiler

10:25 PM - aronpm: let me ask you a question
10:25 PM - aronpm: if you have the first n cubes, and you take away the first (n-1) cubes, what do you have left?
10:26 PM - ***: So you're saying that if you had 50 cubes and took away the first 49 cubes what you'd have left?
10:26 PM - aronpm: yes
10:26 PM - ***: The last cube.
10:26 PM - ***: I think my 6 year old niece could get that one.

10:31 PM - aronpm: btw when I said cubes you realised I meant like 1^3 + 2^3 + ... + n^3 right?
10:31 PM - aronpm: - (1^3 + 2^3 + ... (n-1)^3)
10:31 PM - ***: Yeah, but the clever play on words seemed like a simpler way to explain it.

I mean, he did truncate off the zero from \( \sum\limits_{i=1}^{n}{\left( n-1 \right)^{3}} \)

No, that was me, and I did it for brevity. 0^3 is 0, and I am pretty certain that if you're reading a proof with summations, you know that subtracting 0 doesn't do a damn thing. Are you really going to complain that I left out a 0^3 term? I guess I should start writing \( \int_0^1 \! x^2 \, \mathrm{d}x = \frac{1^3}{3} - \frac{0^3}{3} \) too

but then when you match the terms to subtract, you must add one more term to be able to subtract the first n cubes from each other.

No, you're not subtracting the first n cubes from each other. That's why it explicitly says (i-1) right there. You're not stopping at n^3, you stop at (n-1)^3

\( 1^{3}+2^{3}+\cdot \cdot \cdot +n^{3}-\left( 1^{3}+2^{3}+\cdot \cdot \cdot \left( n-1 \right)^{3} \right)=n^{3}-\left( n-1 \right)^{3}\ne n^{3} \).

You really wanted me to specify that the n is the same in both sums? It's not an index, it doesn't get 'reset' every time I move onto to a different sum. It's a constant variable throughout the entire thing.

I cannot see how you can know that \( \sum\limits_{i=1}^{n}{i^{3}}-\sum\limits_{i=1}^{n}{\left( i-1 \right)^{3}}=n^{3} \).

I cannot see how you can't see that from the expansion (or even saying it in words, re: spoilered conversation)