Lesson The Pascal's triangle

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The Pascal's triangle

This lesson is about the Pascal's triangle. You should know what the binomial coefficients C%5Bn%5D%5Em

are.
You can find all the necessary information in the lesson Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion under the current topic in this site.

Definition. The Pascal's triangle is the triangle of numbers formed by the binomial coefficients    in a way shown in the Figure 1 and the Figure 2 below.







      . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
                   . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Figure 1. The Pascal's triangle in terms of binomial coefficients













      . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
                   . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

      Figure 2. The Pascal's triangle in numerical terms

The upper row contains only one number       = .                                       (By the agreement, this row has the number 0, "zero".)
The next row contains two numbers               = and = .              (This row has the number 1.)
The row number 2 contains three numbers    = , = and = .
The row number 3 contains four numbers      = , = , = and = .
The row 4 contains five numbers                     = , = , = , = and = .
The row 5 contains six numbers                       = , = , = , = , = and = .
And so on ...

The n-th row contains numbers                       = , = , = , = , . . . , = = , . . . ,
                                                                                                               =, = and = .

The major property of the Pascal's triangle

In the Pascal triangle, the very first and the very last
number in each row is equal to 1.

Starting from the row number 2, each number between
the very first and very last is equal to the sum of two
its closest neighbors in the previous row.

In mathematical terms, this means that

C%5Bn%5D%5Ek

for any positive integer number n and all integer
numbers k = 0, 1, 2, . . . , n.

This property is illustrated in the Figure 3.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Figure 3. The property of the Pascal's triangle

Proof
Actually, this identity was just proved in the lesson Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion under the current topic in this site.
For your convenience, here we repeat the proof again.

We have

C%5Bn%5D%5Ek

%28n%2A%28n-1%29%2A+ellipsis+%2A%28n-k%2B1%29%29%2F%281%2A2%2A+ellipsis+%2Ak%29

= (now express fractions as equivalent fractions with the common denominator 1%2A2%2A+ellipsis+%2Ak%2A%28k%2B1%29

)

= (now take the common factor

out of brackets)

= (now rewrite the expression in the very right brackets)

.

We got the required identity. The proof is completed.

Other properties of the Pascal's triangle

1. In the Pascals triangle the sum of all numbers in the n-th row is equal to .
2. In the Pascals triangle the alternate sum of all numbers in the n-th row is equal to 0.

Let us prove the first of the two properties.
The sum of all numbers of the n-th row of the Pascal's triangle is equal to

C%5Bn%5D%5E0

+ . . . +

.

We have to prove that

C%5Bn%5D%5E0

+ . . . +

.

Use the binomial expansion

%281+%2B+x%29%5En

= =

+sum+%28+C%5Bn%5D%5Ei%2Ax%5Ei+%2C+i=0%2C+n+%29+

(see the lesson Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion under the current topic in this site). Substitute

to the last
formula, and you will get the required identity.

This is how it looks for the low order binomial coefficients and the first rows of the Pascal's triangle:

n = 1, C%5B1%5D%5E0

= 1,

= 1,

= 1 + 1 = 2;

n = 2, C%5B2%5D%5E0

= 1,

= 2,

= 1,

= 1 + 2 + 1 = 4;

n = 3, C%5B3%5D%5E0

= 1,

= 3,

= 3,

= 1,

= 1 + 3 + 3 + 1 = 8.

The proof of the second property is also straightforward.
The alternate sum of all numbers of the n-th row of the Pascal's triangle is equal to

C%5Bn%5D%5E0

+ . . . +

+ . . . +

.

We have to prove that

C%5Bn%5D%5E0

+ . . . +

.

Use the same binomial expansion

%281+%2B+x%29%5En

= =

.

Substitute

to the last formula, and you will get the required identity.

This is how it looks for the low order binomial coefficients and the first rows of the Pascal's triangle:

n = 1, C%5B1%5D%5E0

= 1,

= 1 - 1 = 0;

n = 2, C%5B2%5D%5E0

= 1,

= 2,

= 1,

= 1 - 2 + 1 = 0;

n = 3, C%5B3%5D%5E0

= 1,

= 3,

= 1,

= 1 - 3 + 3 - 1 = 0.

There are other identities for the binomial coefficients and the Pascal's triangle elements.
For example,

C%5Bn%5D%5E0

+ . . . +

+ . . . +

+ . . . +

+ . . . +

+ . . . +

+ . . . +

+ . . . +

%28%28%28-1%29%5Ek%29%2F2%5Ek%29%2AC%5Bn%5D%5Ek

+ . . . +

%28%28%28-1%29%5En%29%2F2%5En%29%2AC%5Bn%5D%5En

for any positive integer number n.

You can prove these and similar identities yourself. Simply use the binomial expansion and make an appropriate substitution.

My lessons on Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion in this site are

    - Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion
    - Remarkable identities for Binomial Coefficients
    - The Pascal's triangle                                                                                                                    (this lesson)
    - Solved problems on binomial coefficients
    - Solving equations that include binomial coefficients and numbers of permutations
    - Math circle level problem on binomial coefficients
    - OVERVIEW of lessons on Binomial Expansion, Binomial coefficients and the Pascal's triangle

Use this file/link ALGEBRA-II - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-II.

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