**Pascal’s Triangle: Polynomials and TI-84 Plus**

Pascal's Triangle |

**Powers of 2**

Pascal’s
Triangle holds a great number of properties.
For instance, the sum of each row (R) is a power of two (2^R). The top row is referred to as row 0. Hence:

1 = 2^0

1 + 1 = 2 = 2^1

1 + 2 + 1 = 4 =
2^2

1 + 3 + 3 + 1 =
8 = 2^3

1 + 4 + 6 + 4 +
1 = 16 = 2^4

1 + 5 + 10 + 10
+ 5 + 1 = 32 = 2^5

1 + 6 + 15 + 20
+ 15 + 6 + 1 = 64 = 2^6

and so on…

**Binomial Expansion**

Take a look
what happens when you expand the binomial (x + y)^R:

(x + y)^0 =

**1**
(x + y)^1 =

**1***x +**1***y
(x + y)^2 =

**1***x^2 +**2***x*y +**1***y^2
(x + y)^3 =

**1***x^3 +**3***x^2*y +**3***x*y^2 +**1***y^3
(x + y)^4 =

**1***x^4 +**4***x^3*y +**6***x^2*y^2 +**4***x*y^3 +**1***y^4
(x + y)^5 =

**1***x^5 +**5***x^4*y +**10***x^3*y^2 +**10***x^2*y^3 +**5***x*y^4 +**1***y^5
(x + y)^6 =

**1***x^6 +**6***x^5*y +**15***x^4*y^2 +**20***x^3*y^3 +**15***x^2*y^4 +**6***x*y^5 +**1***y^6
and so on…

Notice the coefficients
(

**in blue**)? They represent rows of the Pascal’s Triangle.**Combinatorics**

The formula for
find the number of combinations of N items from R is:

COMB(R, N) = R
nCr N = R! / ((R – N)! * N!)

If you R stand
for a row of the Pascal’s Triangle and N stand for an entry (starting from 0),
you can get R nCr N from the triangle.

For instance,
for the 4

^{th}row (R = 4 with entries 1, 4, 6, 4, 1), 2^{nd}entry (N = 2, with left most entry designated as 0), Pascal’s Triangle will state that 4 nCr 3 = 4! / (2! * 2!) = 6.
For the 5

^{th}Row (R = 5):
N = 0, 5 nCr 0
= 1

N = 1, 5 nCr 1
= 5

N = 2, 5 nCr 2
= 10

N = 3, 5 nCr 3
= 10

N = 4, 5 nCr 4
= 5

N = 5, 5 nCr 5
= 1

**Sierpinksi Triangle**

One of the
books I got for Christmas is the book “The Magic of Math: Solving for x and
Figuring Out Why” written by Arthur Benjamin.
The book is well written and if you want a good read I recommend this
book. It has something for everyone. One
of things I learned from Benjamin’s book is that if you mark all the odd
numbers, and you take many rows, you get the famous fractal the Sierpinski
Triangle.

Take a look at the
diagram below:

Making the Sierpinski Triangle |

**TI-84 Plus: Generating a Row of Pascal’s Triangle**

A short program
to generate a row of Pascal’s Triangle.
The result is stored in list L6.
The first entry is the 0

^{th}entry.
Input
"ROW:",R

R+1→dim(L₆)

For(K,0,R)

R nCr K→L₆(K+1)

End

Disp "L₆:"

Pause L₆

I think this
is self-explanatory.

Eddie

HAPPY NEW
YEAR!

Source:

Benjamin,
Arthur. The Magic of Math: Solving for x
and Figuring Our Why. Basic Books: New York.
2015

This blog is
property of Edward Shore. 2015.

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