Expand binomials using Pascal's triangle and binomial theorem. Math 30-1 Alberta.
Lesson 7.4 of Probability Distributions in Math 30-1 — Alberta curriculum lessons.
The Binomial Theorem. The binomial theorem gives us a procedure for expanding expressions of the form (a + b)^n without having to multiply everything out by hand repeatedly.
Before we get to the formula, let's see where it comes from by expanding manually:
( a+b)^0 = 1
( a+b)^1 = a + b
( a+b)^2 = a^2 + 2ab + b^2
( a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
( a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4
( a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5
Notice the coefficients of each expansion: 1, then 1\ 1, then 1\ 2\ 1, then 1\ 3\ 3\ 1, and so on. These form a pattern called Pascal's Triangle.
Pascal's Triangle. Every number is the sum of the two numbers directly above it. The edges are always 1.
arrayccccccccccc & & & & & 1 & & & & & & & & & 1 & & 1 & & & & & & & 1 & & 2 & & 1 & & & & & 1 & & 3 & & 3 & & 1 & & & 1 & & 4 & & 6 & & 4 & & 1 & 1 & & 5 & & 10 & & 10 & & 5 & & 1 array
Connection to combinations: Each number in Pascal's Triangle is a combination value. Row n, position r (starting from 0) = nr.
For example, Row 4 is: 40,\ 41,\ 42,\ 43,\ 44 = 1,\ 4,\ 6,\ 4,\ 1
Key Patterns to Know. 1. The exponent on the binomial = the row number = n
2. The number of terms in the expansion is always n + 1
3. Powers of a start at n and decrease by 1 each term
4. Powers of b start at 0 and increase by 1 each term
5. The sum of the exponents on a and b always equals n
6. For subtraction binomials like (a - b)^n, the signs alternate: +, −, +, −, ...
a) First three terms of (a + b)¹⁵. Step 1 — Identify n.
n = 15, so we use Row 15 of Pascal's Triangle. The coefficients are 150,\ 151,\ 152,\
Step 2 — Write the pattern.
Powers of a start at 15 and decrease. Powers of b start at 0 and increase.
Step 3 — Write the first three terms.
t_1 = 150 · a^(15) · b^0 = a^(15)
t_2 = 151 · a^(14) · b^1 = 15a^(14)b
t_3 = 152 · a^(13) · b^2 = 105a^(13)b^2
(a+b)^(15) = a^(15) + 15a^(14)b + 105a^(13)b^2 +
b) First three terms of (x − 3y)¹². Step 1 — Identify a, b, and n.
Here a = x, b = -3y, n = 12.
Important: Since this is a subtraction binomial, b = -3y (negative), so substitute the full negative value into each term.
Step 2 — Write the first three terms.
t_1 = 120 · x^(12) · (-3y)^0 = x^(12)
t_2 = 121 · x^(11) · (-3y)^1 = 12 · x^(11) · (-3y) = -36x^(11)y
t_3 = 122 · x^(10) · (-3y)^2 = 66 · x^(10) · 9y^2 = 594x^(10)y^2
(x-3y)^(12) = x^(12) - 36x^(11)y + 594x^(10)y^2 -
Create a free Studyio account to take the full lesson with voice-over teaching, practice questions and instant feedback.