Lesson 1 of 5
From repeated multiplication to Pascal's triangle
When (a+b) is multiplied repeatedly, like terms merge in a stable triangular pattern.
Core lesson · about 6 minutes · 3 guided parts
A path down Pascal's triangle chooses left or right at every step. The number written at a point counts how many paths reach it—the same choice count that becomes a coefficient in an expansion.
Your learning route
Build the idea one piece at a time
- Build small expansions
- Pascal's triangle
- See coefficients as paths, not magic numbers
Build small expansions
Multiply carefully first, then let the pattern become visible.
(a+b)² means (a+b)(a+b), so every term in one bracket multiplies every term in the other.
After collecting like terms, (a+b)²=a²+2ab+b² and (a+b)³=a³+3a²b+3ab²+b³.
The powers of a decrease while the powers of b increase, and their exponents always sum to n.
Every term in an expansion comes from one choice in each bracket. Writing small choice paths explains why no term may be skipped when multiplying the brackets.
The middle coefficient 2 in (a+b)² appears because ab is produced in two ways: choose a then b, or b then a. Like products are collected after all choices are made.
Every term of (a+b)ⁿ has total degree n because exactly one degree-one factor is selected from each of n brackets. A term with total degree n−1 signals a lost factor.
(a+b)²=a²+2ab+b²Worked example
Expand (x+2)³ by repeated multiplication.
- First, (x+2)²=x²+4x+4.
- Multiply by (x+2).
- This gives x³+2x²+4x²+8x+4x+8.
- Combine like terms: x³+6x²+12x+8.
Answer: x³+6x²+12x+8
Pascal's triangle
Each interior number is the sum of the two numbers leaning above it.
Row n of Pascal's triangle contains C(n,0),C(n,1),…,C(n,n).
The row is symmetric because choosing r objects is equivalent to choosing the n−r objects left out.
The entries sum to 2ⁿ because they count all subsets grouped by subset size.
Pascal's triangle begins and ends each row with 1 because there is one way to choose none and one way to choose all. Interior entries add the two entries above them.
The addition rule behind Pascal's triangle asks whether a chosen group contains one special item. Those two cases create C(n−1,r−1)+C(n−1,r).
Row checks catch arithmetic errors: row n must be symmetric and sum to 2ⁿ. For row 4, 1+4+6+4+1=16=2⁴.
C(n,r)=C(n−1,r−1)+C(n−1,r)Worked example
Construct row 5 of Pascal's triangle from row 4: 1,4,6,4,1.
- Place 1 at each end.
- Add adjacent entries: 1+4=5.
- Continue: 4+6=10, 6+4=10, 4+1=5.
- The row is 1,5,10,10,5,1.
Answer: 1, 5, 10, 10, 5, 1
Practical deep dive for this lesson
See coefficients as paths, not magic numbers
Pascal's triangle appears because every route to a position ends with one of two final choices, so neighbouring path counts add.
Expanding (a+b)ⁿ means choosing either a or b from each of n identical-looking factors and multiplying the choices.
Terms with the same numbers of a and b become like terms. Their coefficient counts how many different factor-choice paths produced that same product.
In Pascal's triangle, a path moves down-left or down-right. Every path reaching an interior entry must arrive from exactly one of the two entries above it.
That last-step split explains the addition rule C(n,r)=C(n−1,r−1)+C(n−1,r), rather than asking students to memorize a pattern without a reason.
The row is symmetric because choosing r positions for b determines the n−r positions for a. Reading from either end counts the same choices.
The row sum is 2ⁿ because every one of n factors offers two choices. Pascal's entries simply group those 2ⁿ paths by how many b choices occurred.
This same structure counts binary strings, subsets, left-right paths, and success counts, which is why Pascal's triangle reappears throughout probability and computing.
Building small rows by hand is a check, but for a distant row use the combination formula to avoid carrying an early addition error downward.
C(n,r)=C(n−1,r−1)+C(n−1,r)Worked example
Explain and build row 6 of Pascal's triangle, then use it to expand (x+y)⁶.
- Start from row 5: 1,5,10,10,5,1.
- Place 1 at both ends because there is one all-x choice and one all-y choice.
- Add adjacent entries: 1+5=6, 5+10=15, 10+10=20, then symmetry gives 15 and 6.
- Row 6 is 1,6,15,20,15,6,1.
- Pair these coefficients with x powers descending 6,5,…,0.
- Pair y powers ascending 0,1,…,6, keeping each exponent sum equal to 6.
- This gives x⁶+6x⁵y+15x⁴y²+20x³y³+15x²y⁴+6xy⁵+y⁶.
- Check coefficients: their sum is 64=2⁶, the total number of factor-choice paths.
Answer: Row 6 is 1,6,15,20,15,6,1, giving x⁶+6x⁵y+15x⁴y²+20x³y³+15x²y⁴+6xy⁵+y⁶.
Check your picture
What is the coefficient of x² in (1+x)⁴?
Reveal answer
6
This lesson versus the whole chapter
Your topic-specific practical example is above.
The final lesson collects five longer chapter-wide workflows for machine learning, LLMs, trading, physics, and everyday decisions.
Formula shelf
Symbols with meaning
(a+b)ⁿ=Σ C(n,r)aⁿ⁻ʳbʳExpands a power of two terms.
Tᵣ₊₁=C(n,r)aⁿ⁻ʳbʳJumps directly to one term.
C(n,r)=C(n−1,r−1)+C(n−1,r)Builds each interior Pascal entry.
ΣC(n,r)=2ⁿCounts all subsets of n objects.
P(X=r)=C(n,r)pʳ(1−p)ⁿ⁻ʳProbability of exactly r successes.
E[X]=np, Var(X)=np(1−p)Gives centre and spread.
n!/(n₁!⋯nₖ!)Counts repeated choices across many categories.
Before moving on
You are ready when you can…
- Build and read Pascal's triangle and explain its counting meaning.
- Expand positive-integer binomial powers with correct coefficients, signs, and exponents.
- Find a requested term, power, or coefficient without full expansion.
- Use coefficient identities and solve binomial probability questions.
- Apply truncated generalized expansions and multinomial coefficients with their conditions.