Need the coefficients of (a+b)n(a+b)^n without multiplying the whole thing out? Pascal's triangle generates them with one local rule: each interior entry is the sum of the two entries directly above it, and the edges are all 11. Counting the top as row 00, row nn holds the coefficients of (a+b)n(a+b)^n.

When to use this method

Reach for the triangle when you want binomial coefficients, a quick binomial expansion, simple combination counts, or a check on an expansion you got another way. It is fastest for small to moderate nn, where building rows by hand beats computing each (nk)\binom{n}{k} separately. The first few rows:

Row 0:1Row 1:1, 1Row 2:1, 2, 1Row 3:1, 3, 3, 1Row 4:1, 4, 6, 4, 1\begin{aligned} &\text{Row } 0: \quad 1\\ &\text{Row } 1: \quad 1,\ 1\\ &\text{Row } 2: \quad 1,\ 2,\ 1\\ &\text{Row } 3: \quad 1,\ 3,\ 3,\ 1\\ &\text{Row } 4: \quad 1,\ 4,\ 6,\ 4,\ 1 \end{aligned}

The steps

  1. Start with 1. Place a single 11 at the top.
  2. Keep 1s on the edges. Every new row begins and ends with 11.
  3. Add the two numbers above. Each interior entry is the sum of its two upper neighbors.
  4. Read the pattern. With the top as row 00, row nn gives the coefficients for (a+b)n(a+b)^n.

For example, in row 44 the middle 66 comes from 3+33 + 3, and the 44 beside it from 1+31 + 3 — so each row is built straight from the row before it, no separate formula needed.

Why a row equals the binomial coefficients

Row nn can also be written with combinations:

(n0), (n1), (n2), , (nn)\binom{n}{0},\ \binom{n}{1},\ \binom{n}{2},\ \dots,\ \binom{n}{n}

where (nk)\binom{n}{k} counts the ways to choose kk objects from nn. That is why row 44,

1, 4, 6, 4, 11,\ 4,\ 6,\ 4,\ 1

gives

(a+b)4=a4+4a3b+6a2b2+4ab3+b4,(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4,

and why the triangle ties algebra to counting.

The whole procedure on one example

Expand (x+y)5(x+y)^5. With the top as row 00, row 55 is

1, 5, 10, 10, 5, 11,\ 5,\ 10,\ 10,\ 5,\ 1

Match those coefficients with descending powers of xx and ascending powers of yy:

(x+y)5=x5+5x4y+10x3y2+10x2y3+5xy4+y5(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5

The triangle supplies the coefficients, but you still place the powers in order: the exponent of xx falls from 55 to 00 while the exponent of yy rises from 00 to 55.

Where each step tends to stall, and how to check

Step 3 (adding above): only the two entries directly above a position count, never numbers off to the side. Step 4 (row numbering): if a source starts rows at 11 instead of 00, the coefficient row for (a+b)n(a+b)^n is labeled differently, so fix your convention first. Placing powers: the triangle alone is not the full expansion — write the powers correctly too.

Two properties give you fast self-checks. Each row is symmetric, since (nk)=(nnk)\binom{n}{k} = \binom{n}{n-k}, so row 55 reads 1,5,10,10,5,11,5,10,10,5,1 the same both directions. And the row sum is 2n2^n:

1+5+10+10+5+1=32=251+5+10+10+5+1 = 32 = 2^5

If your row 55 does not add to 3232, something went wrong.

Build row 66 from row 55, then use it to expand (m+n)6(m+n)^6 — a clean way to drill both halves of the skill: generating the coefficients and placing the powers. In school math, Pascal's triangle usually appears right before or alongside the binomial theorem, and it stays useful as a quick coefficient check long after.

Frequently Asked Questions

What is Pascal's triangle?
Pascal's triangle is a triangular arrangement of numbers where each interior entry is the sum of the two entries directly above it. The outside edges are all $1$.
How does Pascal's triangle connect to binomial expansion?
If you count the top as row $0$, then the entries in row $n$ are the coefficients in the expansion of $(a+b)^n$.
Why do the rows look symmetric?
They are symmetric because the binomial coefficients satisfy $\binom{n}{k} = \binom{n}{n-k}$, so the numbers match from left to right.

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