The Binomial Theorem
The Binomial Theorem expands \((a+b)^n\) without multiplying it out term by term: \((a+b)^n = \sum_{k=0}^{n} C(n,k)\cdot a^{n-k}\cdot b^k\), where each binomial coefficient \(C(n,k)\) is exactly an entry of Pascal's triangle.
By the end you'll be able to read a row of Pascal's triangle straight off the diagram and use it to expand \((a+b)^n\) term by term, without computing a single factorial.
Predict: what's the coefficient of the \(x^2\) term in \((x+1)^4\)? Set row n to 4 below and read the third entry of that row to check.
Pick a row n of Pascal's triangle below — that row lights up, and the expansion of \((x+1)^n\) assembles underneath, one term per entry. Each term's color matches the triangle entry it came from; hover or focus either one to see the link.
Row 4: 1, 4, 6, 4, 1 → \((x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1\)
The Binomial Theorem turns the tedious work of multiplying out \((a+b)^n\) into reading coefficients off a triangle and pairing them with a predictable pattern of exponents.
\((a+b)^n = \sum_{k=0}^{n} C(n,k)\cdot a^{n-k}\cdot b^k\), where \(C(n,k) = n!/(k!(n-k)!)\). As k runs from 0 to n, the exponent on a falls from n to 0 while the exponent on b rises from 0 to n — the two exponents always add up to n in every term. This is the same sum written compactly with sigma notation, one of the tools you've already met for writing series.
To expand \((a+b)^n\): (1) read row n of Pascal's triangle for the coefficients, (2) write the a-exponents counting down from n to 0, (3) write the b-exponents counting up from 0 to n, matching position for position, and (4) attach each coefficient to its pair of exponents and add the terms together.
Binomial coefficients aren't just an algebra shortcut — they're also a counting tool. \(C(n,k)\) answers "in how many ways can I choose k items from a set of n?", which is exactly why the coefficient of the term with \(b^k\) counts the number of ways k of the n factors of \((a+b)(a+b)\cdots(a+b)\) can each contribute a b (and the rest contribute an a).
Expand \((a+b)^4\). Row 4 of Pascal's triangle: 1, 4, 6, 4, 1. Pair each with falling/rising exponents: \(a^4b^0, a^3b^1, a^2b^2, a^1b^3, a^0b^4\). So \((a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\). Check the middle term with the formula: \(C(4,2) = 4!/(2!\cdot 2!) = 6\), matching the coefficient on \(a^2b^2\). ✓
Expand \((x+2)^3\) — here \(a = x\) and \(b = 2\), so the b's carry a value into the arithmetic. Row 3: 1, 3, 3, 1. \(k=0\): \(1\cdot x^3\cdot 2^0 = x^3\). \(k=1\): \(3\cdot x^2\cdot 2^1 = 6x^2\). \(k=2\): \(3\cdot x^1\cdot 2^2 = \) ____. \(k=3\): \(1\cdot x^0\cdot 2^3 = 8\).
Reveal the answer
\(k=2\): \(3\cdot x\cdot 4 = \) 12x, so \((x+2)^3 = x^3 + 6x^2 + 12x + 8\). Set row n to 3 on the slider above (mentally swap in \(b=2\)) and check the coefficients 1, 3, 3, 1 against the triangle.
More info — where Pascal's triangle rule comes from
Each entry of Pascal's triangle is the sum of the two entries diagonally above it: \(C(n,k) = C(n-1,k-1) + C(n-1,k)\). That's not a coincidence — a term with \(b^k\) in \((a+b)^n\) either came from a term with \(b^{k-1}\) in \((a+b)^{n-1}\) that just picked up one more b, or from a term that already had \(b^k\) and picked up an a. Adding those two counts gives the addition rule that builds every row from the one above it. See the Math is Fun link in Dive deeper for more worked rows.
Check your understanding
Using the Binomial Theorem, what is the coefficient of the \(a^4b^2\) term in \((a+b)^6\)?
Why is \(C(n,k)\) always equal to \(C(n,n-k)\)?
For any row n of Pascal's triangle, what are the first and last entries, \(C(n,0)\) and \(C(n,n)\)?
Which sigma-notation sum correctly matches the expansion of \((a+b)^n\)?
Recap
- \((a+b)^n = \sum_{k=0}^{n} C(n,k)\cdot a^{n-k}\cdot b^k\); a's exponent falls from n to 0 while b's rises from 0 to n.
- \(C(n,k) = n!/(k!(n-k)!)\) — row n of Pascal's triangle lists \(C(n,0)\ldots C(n,n)\).
- Every row starts and ends in 1 (\(C(n,0)=C(n,n)=1\)), and \(C(n,k)=C(n,n-k)\) — rows read the same forwards and backwards.
- Each entry is the sum of the two above it: \(C(n,k) = C(n-1,k-1)+C(n-1,k)\), which is how the triangle grows one row at a time.
Dive deeper
- Math is Fun — Binomial Theorem Assemble an expansion term by term.
- Math is Fun — Pascal's Triangle Read binomial coefficients straight off the rows.
Sources
- The Binomial Theorem and Pascal's Triangle