Is ##\sum^n_{k=0} 2k+1 = n^2## useful? Has it been found already?

[MevsEinstein]

TL;DR

What the title says

I was looking at the tiles of my home's kitchen when I realized that you can form squares by summing consecutive odd numbers. First, start with one tile, then add one tile to the right, bottom, and right hand corner (3), and so on. Can this be applied somewhere? And has someone found it already?

 

[FactChecker]

This is an example of a family of summations where you can add the sum to the reversed sum and get a simple solution. I don't get exactly the same answer as you got. (I assumed that the 1 is outside of the summation.)
Consider this:

2* summation
=  summation
  + summation reversed
=   0 +        2 +         4  + ...+(2n-4)+(2n-2)+2n
  +2n +    (2n-2)+     (2n-4) + ... +    4 +    2 + 0
 = (n+1) *(2n)

So summation = n^2 + n.

 

[MevsEinstein]

This is an example of a family of summations where you can add the sum to the reversed sum and get a simple solution. I don't get exactly the same answer as you got. (I assumed that the 1 is outside of the summation.)
Consider this:

2* summation
=  summation
  + summation reversed
=   0 +        2 +         4  + ...+(2n-4)+(2n-2)+2n
  +2n +    (2n-2)+     (2n-4) + ... +    4 +    2 + 0
 = (n+1) *(2n)

So summation = n^2 + n.

your summation goes from $k=1$ to $n$, but I began from $k=0$. When I was searching my sum in Google, I saw the sum you described, but they didn't include 1.

 

[Office_Shredder]

Your equation isn't quite right I think. When n=1 you get 1+3=1 which is obviously wrong. The right hand side should be $(n+1)^2$

This is a very well known result. I'm sure it's useful in a variety of applications.

 

[Svein]

https://en.wikipedia.org/wiki/Arithmetic_progression

 

[PeroK]

Summary: What the title says

I was looking at the tiles of my home's kitchen when I realized that you can form squares by summing consecutive odd numbers. First, start with one tile, then add one tile to the right, bottom, and right hand corner (3), and so on. Can this be applied somewhere? And has someone found it already?

Yes, it's well known. You can use it to prove that there are infinitely many distinct Pythagorean triples:

Let $k$ be any odd number. Note that $k^2$ is odd, hence the difference between two consecutive squares. I.e. we can find $n$ such that $n^2 + k^2 = (n+1)^2$.

That means we have a Pythagorean triple $(k, n, n+1)$. And, as in general $n$ and $n+1$ have no common factors, it cannot reduce to a lower form.

Can you find a similar pattern for consecutive cubes?

 

[FactChecker]

your summation goes from $k=1$ to $n$,

I don't understand why you say that. When k=0, 2k=0. That is the first term in my post.

 

[MevsEinstein]

I don't understand why you say that. When k=0, 2k=0. That is the first term in my post.

Oh I thought the first term for you was 3.

 

[MevsEinstein]

Can you find a similar pattern for consecutive cubes?

1=1, 1+8=9, 1+8+27=36, 1+8+27+64=100, ... In other words, this is $1^2, 3^2, 6^2, 10^2$. 1 is the sum of the first integer, 3 is the sum of the first two consecutive integers, and so on. this means that the sum of consecutive cubes is $\frac{n^2*(n+1)^2}{4}$. I have seen this before.

 

[MevsEinstein]

Can you find a similar pattern for consecutive cubes?

Oh wait, I misunderstood what you were saying. So we have 1, 7, 19, 37... To me this looks like a quadratic equation. After solving some systems of equations, we now know that $\sum_{k=0}^n 3k^2 - 3k +1 = n^3$. Amazing!

Discovering math on my own is feeling great :).

 

[PeroK]

Oh wait, I misunderstood what you were saying. So we have 1, 7, 19, 37... To me this looks like a quadratic equation. After solving some systems of equations, we now know that $\sum_{k=0}^n 3k^2 - 3k +1 = n^3$. Amazing!

Discovering math on my own is feeling great :).

Let's take the sequence of differences between consecutive cubes: $1, 7, 19, 37 \dots$.

Can you guess the next one from that sequence?

 

[MevsEinstein]

Let's take the sequence of differences between consecutive cubes: $1, 7, 19, 37 \dots$.

Can you guess the next one from that sequence?

the pattern seems to be the consecutive multiples of six, so I think the next term is 61.

 

[MevsEinstein]

$\sum_{k=0}^n 3k^2 - 3k +1 = n^3$. Amazing!

$k$ actually goes from $1$ to $n$, sorry about that.