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Being naturally curious, i was thinking by myself whether there might not be a way to determine the sum of all the numbers in this triangle by some method, depending on the number of rows it has. And also the sum of the numbers in the n-th row.
This is what i've discovered so far.
Row 1 = 1 ; Sum = 1
Row 2 = 1 1 ; Sum = 2
Row 3 = 1 2 1 ; Sum = 4
Row 4 = 1 3 3 1 ; Sum = 8
Row 5 = 1 4 6 4 1 ; Sum = 16
So the sum of these numbers in each row, are multiples of 2.
seems to be $\displaystyle 2^{n-1}$, where $\displaystyle n$ is the number of the row we are at from what you have thereQuote:
Originally Posted by janvdl
Ah ok, i wonder why i didnt notice that. And the sum of all the numbers up to an n-th row? :)Quote:
Originally Posted by Jhevon
seems to be $\displaystyle 2^n - 1$ where $\displaystyle n$ is the number of the row we are atQuote:
Originally Posted by janvdl
I was thinking in that line. :) cant believe i was so blind :DQuote:
Originally Posted by Jhevon
Hello, janvdl!
Good for you!Quote:
Being naturally curious . . .
I made some amazing "discoveries" while exploring on my own.
Of course, anything I found was proved centuries ago,
. . but it was still very satisfying.
What you discovered is a stunning pattern.
Consider the coefficients of the expansion of $\displaystyle (a + b)^n$
. . . $\displaystyle \begin{array}{ccc}n & \text{coefficients} & \text{sum} \\ \hline 0 & 1 & 1 = 2^0\\ 1 & 1\;\;1 & 2 = 2^1 \\ 2 & 1\;\;2\;\;1 & 4 = 2^2 \\ 3 & 1\;\;3\;\;3\;\;1 & 8 =2^3 \\ 4 & 1\;\;4\;\;6\;\;4\;\;1 & 16 = 2^4 \\ 5 & 1\;\;5\;\;10\;\;10\;\;5\;\;1 & 32 = 2^5 \\ 6 & 1\;\;6\;\;15\;\;20\;\;15\;\;6\;\;1 & 64 = 2^6 \\ \vdots & \vdots & \vdots \end{array}$
The sum of the $\displaystyle n^{th}$ row is $\displaystyle 2^n$.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I "discovered" this pattern while in college.
$\displaystyle \begin{array}{ccc}1 + 2 & = & 3 \\ 4 + 5 + 6 & = & 7 + 8 \\ 9 + 10 + 11 + 12 & = & 13 + 14 + 15 \\ 16 + 17 + 18 + 19 + 20 & = & 21 + 22 + 23 + 24 \\
\vdots & & \vdots\end{array}$
I found these in a book . . .
$\displaystyle \begin{array}{ccc}1 & = & 1 = 1^3 \\ 3 + 5 & = & 8 = 2^3 \\ 7 + 9 + 11 & = & 27 = 3^3 \\ 13 + 15 + 17 + 19 & = & 64 = 4^3 \\ 21 + 23 + 25 + 27 + 29 & = & 125 = 5^3 \\ \vdots & & \vdots \end{array}$
$\displaystyle \begin{array}{ccc}3^2 + 4^2 & = & 5^2 \\ 10^2 + 11^2 + 12^2 & = &13^2 + 14^2 \\ 21^2 + 22^2 + 23^2 + 24^2 & = & 25^2 + 26^2 + 27^2 \\ 36^2 + 37^2 + 38^2 + 39^2 + 40^2 & = & 41^2 + 42^2 + 43^2 + 44^2 \\ 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 & = & 61^2 + 62^2 + 63^2 + 64^2 + 65^2 \\ \vdots & &\vdots\end{array}$
I've found something else too...
I was actually trying to find another type of formula when i noticed this...
I attached a picture, i know its badly drawn but at least it gives you the idea...
The white triangles are the normal figures of Pascal's Triangle. But look at those red triangles. They interest me a lot. There must be a lot more patterns involved in this triangle than i ever thought.
What if every red triangle is the product of the 3 white triangles around it? Or the sum of the 3 white ones around it? We can make lots of new patterns using this...
Hello, janvdl!
That's fascinating . . . Thank you!
Quote:
What if every red triangle is the sum of the 3 white ones around it?
Then we have:
. . $\displaystyle \begin{array}{c}3\\
4\;\;4\\
5\;\;8\;\;5 \\
6\;\;13\;\;13\;\;6 \\
7\;\;19\;\;26\;\;19\;\;7
\end{array}$
I finally saw where it came from . . .
Consider a "Pascal's Triangle" whose third line is: .1 - 3 - 1.
. . $\displaystyle \begin{array}{c}1 \\ 1\;\;1 \\ 1\;\;3\;\;1 \\
1\;\;4\;\;4\;\;1 \\ 1\;\;5\;\;8\;\;5\;\;1 \\ 1\;\;6\;\;13\;\;13\;\;6\;\;1 \\ 1\;\;7\;\;19\;\;26\;\;19\;\;7\;\;1
\end{array}$
See it?
This opens up a world of possibilities, doesn't it?
So have i actually kind of discovered something here? :DQuote:
Originally Posted by Soroban
And what can we actually do with my "discovery"? A world of possibilities has been opened? You're clearly seeing things that i cant... :D
Here's another of my "discoveries", made years and years ago.
We know the Binomial Theorem is based on Pascal's Triangle.
We use Pascal's Triangle for expanding $\displaystyle (a + b)^n$.
I wondered if there was a "Trinomial Theorem" for expanding $\displaystyle (a + b + c)^n$.
I cranked out the first few cases . . . and got into an awful mess.
. . $\displaystyle \begin{array}{ccc}(a + b + c)^0 & = & 1 \\
(a + b + c)^1 & = & a + b + c \\
(a + b + c)^2 & = & a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \\
(a + b + c)^3 & = & a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 + 6abc
\end{array}$
How do we make sense out of these terms and coefficients?
It took me the longest time to see what to do:
. . arrange the terms in triangular arrays.
$\displaystyle \begin{array}{ccccc} & & & & a^4 \\ & & &a^3 & 4a^3b\;\;4a^3c \\& & a^2 & 3a^2b\;\;3a^2c & 6a^2b^2\;\;12a^2bc\;\;6a^2c^2 \\& a & 2ab\;\;2ac & 3ab^2\;\;6abc\;\;3ac^2 & 4ab^3\;\;12ab^2c\;\;12abc^2\;\;4ac^3 \\1\quad &\;\; b\;\;c\quad & b^2\;\;2ab\;\;c^2\quad & b^3\quad3b^2c\quad3bc^2\quad c^3\quad & b^4\quad4b^3c\quad6b^2c^2\quad 4bc^3\quad c^4\end{array}$
Look at the coefficients:
$\displaystyle \begin{array}{ccccc} & & & & 1 \\& & & 1 & 4\;\;4 \\& & 1 & 3\;\;3 & 6\;\;12\;\;6 \\& 1 & 2\;\;2 & 3\;\;6\;\;3 & 4\;\;12\;\;12\;\;4 \\1\quad & \;\;\;1\;\;1\quad & \;\;1\;\;2\;\;1\quad & \;\;1\;\;3\;\;3\;\;1\quad & 1\;\;\;4\;\;\;6\;\;\;4\;\;\;1\end{array}$
If we "stack" these triangles, we get a tetrahedron.
. . Each number is the sum of the three numbers above it.
I seriously doubt that I'm the first to notice all this.
But you can feel free to call it "Soroban's Tetrahedron".
I'm just going to try and get some formulas out of my triangle :) Im busy writing a small book on my discoveries in this triangle. Just for the fun of it :D
