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

In summary: Also, I don't think that equation is quite right. The correct one should be ##\sum_{k=0}^n 3k^2 + 3k + 1 = (n+1)^3##.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,
  • #1
MevsEinstein
124
36
TL;DR 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?
 
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  • #2
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:
Code:
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.
 
  • #3
FactChecker said:
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:
Code:
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.
 
  • #4
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.
 
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  • #6
MevsEinstein said:
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?
 
  • #7
MevsEinstein said:
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.
 
  • #8
FactChecker said:
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.
 
  • #9
PeroK said:
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.
 
Last edited:
  • #10
PeroK said:
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 :).
 
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  • #11
MevsEinstein said:
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?
 
  • #12
PeroK said:
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.
 
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  • #13
MevsEinstein said:
##\sum_{k=0}^n 3k^2 - 3k +1 = n^3##. Amazing!
##k## actually goes from ##1## to ##n##, sorry about that.
 

1. Is the equation ##\sum^n_{k=0} 2k+1 = n^2## commonly used in scientific research?

Yes, the equation ##\sum^n_{k=0} 2k+1 = n^2## is commonly used in various fields of science, such as mathematics, physics, and computer science. It is often used to calculate the sum of odd numbers up to a given value of n, and has many applications in different areas of research.

2. How is the equation ##\sum^n_{k=0} 2k+1 = n^2## derived?

The equation ##\sum^n_{k=0} 2k+1 = n^2## can be derived using various mathematical techniques, such as algebraic manipulation, induction, and geometric reasoning. It is a well-known identity that has been proven by many mathematicians throughout history.

3. Can the equation ##\sum^n_{k=0} 2k+1 = n^2## be used in real-life applications?

Yes, the equation ##\sum^n_{k=0} 2k+1 = n^2## has many real-life applications, such as in calculating the total number of objects in a triangular arrangement or in determining the number of squares on a chessboard. It is also used in computer algorithms and programming to solve various problems.

4. Has the equation ##\sum^n_{k=0} 2k+1 = n^2## been discovered or used before?

Yes, the equation ##\sum^n_{k=0} 2k+1 = n^2## has been known and used for centuries by mathematicians and scientists. It has been used in various forms in ancient civilizations, such as in the works of Greek mathematicians like Pythagoras and Archimedes.

5. Are there any variations of the equation ##\sum^n_{k=0} 2k+1 = n^2##?

Yes, there are many variations of the equation ##\sum^n_{k=0} 2k+1 = n^2##, depending on the starting and ending values of k. For example, the equation ##\sum^n_{k=1} 2k+1 = (n+1)^2## starts with k=1 instead of k=0. These variations have their own unique applications and uses in different fields of science.

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