The Fibonacci Series is found in Pascal’s Triangle.
Pascal’s Triangle, developed by the French Mathematician Blaise Pascal, is formed by starting with an apex of 1. Every number below in the triangle is the sum of the two numbers diagonally above it to the left and the right, with positions outside the triangle counting as zero.
The numbers on diagonals of the triangle add to the Fibonacci series, as shown below.
Pascal’s triangle has many unusual properties and a variety of uses:

Horizontal rows add to powers of 2 (i.e., 1, 2, 4, 8, 16, etc.)

The horizontal rows represent powers of 11 (1, 11, 121, 1331, 14641) for the first 5 rows, in which the numbers have only a single digit.

Adding any two successive numbers in the diagonal 13610152128… results in a perfect square (1, 4, 9, 16, etc.)

It can be used to find combinations in probability problems (if, for instance, you pick any two of five items, the number of possible combinations is 10, found by looking in the second place of the fifth row. Do not count the 1’s.)

When the first number to the right of the 1 in any row is a prime number, all numbers in that row are divisible by that prime number
rr says
thnx
babu sasi says
what is this? i want it’s construction.
Gary Meisner says
The illustration above shows how the numbers on the diagonals of Pascal’s triangle add to the numbers of the Fibonacci series. What other type of construction do you seek? Perhaps you can find what you seek at Pascal’s Triangle at Wikipedia.
Erika says
Wonderful video. I love approaching art and degisn from a maths and scientific angle and this illustrates that way of working perfectly. Plus, I only just noticed the link to further explanations so it’s even more exciting.Great post.
Monica says
the exterior of the triangle is made up of 1’s and the rest of the numbers are each the sum of their neighbours from the row above them. 2=1+1, 4=3+1, 21=6+15, etc.
Anastacia Reynolds says
Ohhhhh. Now I get it! Thank you soo much!
Dylan says
This is used for algebra
harvey says
no its not
:/
Heather says
Uh, yes it is Harvey. One common use is for binomial expansion.
David says
Yes, it is. As Heather points out, in binomial expansion.
For instance (X+Y)^4 = 1 XXXX + 4 XXXY + 6 XXYY + 4XYYY + 1YYYY
where the coefficients ( 1, 4, 6, 4, 1 ) are the fourth row of Pascal’s Triangle.
Duhan says
Hey that is very helpful and all but what is the formula to work out the triangle?
Gary Meisner says
Every number in the triangle is the sum of the two numbers diagonally above it to the left and the right, with positions outside the triangle counting as zero.
Duhan says
Thanks
Carl Balanon says
This is good source of information. I used to get ideas from here. Is pascal’s triangle found in fibonacci sequence?
Cole says
yes it does on the shallow diagonals
george says
(a+b)^7 solve
joe says
a^7+a^6*b+a^5*b^2+a^4*b^3+a^3*b^4+a^2*b^5+a*b^6+b^7
Heather says
Almost correct, Joe. Remember to include the coefficients.
H says
That’s where Pascal’s triangle comes in… so (a+b)^7 = 1*a^7 + 7*a^6*b + 21*a^5*b^2 + 35*a^4*b^3 + 35*a^3*b^4 + 21*a^2*b^5 + 7*a*b^6 + 1*b^7.
delrio says
hello, so good information about maths
dany says
Thanks this helped SOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO MUCH.
Vikrant says
Hi,
Can you explain how Pascal’s triangle works for getting the 9th & 10th power of 11 and beyond?
Thanks,
Vikrant
Parviz says
if you see each horizontal row as one number (1,11,121,1331 etc.) it will show the powers of 11 just carry on the triangle and you should be able to find whatever power of 11 your looking for
M.D. says
Carry over the tens, hundreds etc so 1 5 10 10 5 1 becomes 161051 and 1 6 15 20 15 6 1 becomes 1771561.
1
…5
…1 0
…….1 0
…………5
…………….1
___________+
1 6 1 5 1
Daniel Baldock says
I agree with kyle v.i.h ftw.
qwertyuiop says
Interesting
Mark says
You can represent the triangle as a square. Rows & columns represent the decimal expension of powers of 1/9 (= o.111111 ; 1/81 = 0,0123456 ; 1/729 = 0.00136.)
Heather says
This is such an awesome connection. I hadn’t seen that before. Thanks for the visual!
N says
Hi, just wondering what the general expression for Tn would be for the fibonacci numbers in pascal’s triangle? Thanks
Hayley says
This is so useful thanks so so so so so much 😉
john says
the 2nd statement is not at all true, The horizontal rows represent powers of 11 (1, 11, 121, 1331, 14641, 1621051!=.15101051, etc…)
only works for the first 5 rows
11^0=1
11^1=11
11^2=121
11^3=1331
11^4=14641
11^5=161051 is different than 15101051
Gary Meisner says
Good observation. Correction made to the text above. Thanks.
George Frank says
Finding your presentation and explanation of Pascal’s Triangle was very interesting and its analysis amusing.
What is remarkable is to find how each number fits in perfect order inside the triangular matrix to produce all
those amazing mathematical relationships. Thank you so much..!!!
Ayan Shah says
A bit of modification in the horizontal representation resulting in powers of 11 can turn it into a general formula for any power . It goes like this Instead of choosing the numbers directly from the triangle we think each number as a part of a decimal expansion i.e. 1 2 1 =(1 x 100) +(2 x 10) + (1 x 1) . = 11^2 . Similarly it works even for powers greater than 5, for example : 1 6 15 20 15 6 1 = 11^6….. and so on 😉
Sarah says
do you think this is pascal’s ?
1
4 9
16 25 36
49 64 81 100
121 144 169 196 225
Elijah says
You can also find sierpinski’s triangle by marking all odd numbers
Mark says
Althought known as Pascal’s triangle, apparently Pascal himself wrote it as a square. As a square rows and columns represent negative powers of 9 (101).
1 1 1 1 1 1
1 2 3 4 5
1 3 6 10
1 4 10
1 5
1
1/9 = 0,1111111
1/81=0,0123456
1/729= 0.00137
etc.
(using 1/99…. will avoid carrying over of decimals)
Addiing up those fractions ‘aproaches’ the ratio 1/8 = 0,125 (0,1249999999…..)
Similar the infinite sum of negative powers of 90 (1/90) results in 1/89, which decimally represents the diagonal sum of Pascal’s triangle:
1 1 1 1 1 …
0 0 1 2 3 4 …
0 0 0 0 1 3 6 …
0 0 0 0 0 0 1 4 …
0 0 0 0 0 0 0 0 1 …
—————————— +
1 1 2 3 5 …
Mark says
Another application:
(1x) 21 =
(1x) 8 + (1x) 13 =
(1x) 3 + (2x) 5 + (1x) 8 =
(1x) 1 + (3x) 2 + (3x) 3 + (1x) 5 =
(1x) 0 + (4x) 1 + (6x) 1 + (4x) 2, (1x) 3 = 21