Binomial Theorem

The binomial theorem expands a plus b to any power without multiplying it out, with coefficients that count the ways to choose each mix of terms.

Pascal row 1, 3, 3, 1 expands a plus b cubed term by term, and setting both to 1 checks the coefficients sum to 8

Why Does This Exist?

Multiplying $(a + b)^{10}$ out by hand means 1024 micro-products before collecting like terms: slow, error-prone, and pointless, because the pattern is fully known. The binomial theorem writes the answer directly: each coefficient counts how many micro-products produce that term. It converts exponential mechanical work into one counting lookup per term.

It is the backbone of anything with two outcomes repeated: coin flips, click versus skip, success counts in $n$ trials. The binomial distribution is this theorem normalized into probabilities, and polynomial feature expansions lean on the same coefficients. Counting foundations are in combinatorics and counting; the probability sequel is probability distributions.

The one idea you need first

" $n$ choose $k$ ," written $\binom{n}{k}$ , counts the ways to pick $k$ items from $n$ : $\binom{n}{k} = n!/(k!(n-k)!)$ . Expanding $(a+b)^n$ creates one micro-product per $n$ -letter word in $a$ and $b$ ; the term $a^{n-k}b^k$ collects exactly the words with $k$ copies of $b$ . Its coefficient must be $\binom{n}{k}$ : that is the whole theorem in one sentence.

Think of It Like This

Three coin flips and the scoreboard

Flip a coin three times and list all 8 outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Group by number of tails: one outcome with none, three with one, three with two, one with three. The group sizes $1, 3, 3, 1$ are the expansion coefficients: each counts the routes to that mix.

Pascal's triangle is the scoreboard across flip counts: each entry sums the two above because each route to this mix extends a route from the previous round. The analogy stops at signs: with $(a - b)$ the counts still hold but terms alternate sign, which coins cannot show.

How It Actually Works

For non-negative integer $n$ :

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k}\, a^{n-k} b^k

In plain words: the sum runs over $k$ , the number of $b$ factors; $\binom{n}{k}$ counts the micro-products with that mix; $a^{n-k}b^k$ is the term they share. The coefficients of row $n$ of Pascal's triangle are exactly these counts, each the sum of the two above it: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ .

Two instant checks: setting $a = b = 1$ gives $\sum_k \binom{n}{k} = 2^n$ (all $2^n$ outcomes counted); setting $a = 1, b = -1$ gives alternating sum $0$ for $n \geq 1$ (cancellation). Any expansion failing these is wrong.

Worked example

Expand $(a + b)^3$ with row $1, 3, 3, 1$ :

(a+b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3

Verify three ways. At $a = b = 1$ : $1 + 3 + 3 + 1 = 8 = 2^3$ . At $a = 2, b = 1$ : $8 + 12 + 6 + 1 = 27 = 3^3$ . And $\binom{3}{1} = 3!/(1!2!) = 3$ confirms the middle coefficient by direct counting. Expansion, powers, and combinatorics all agree.

Code

Expand symbolically by counting, then verify numerically.

1import math2
3n = 34coeffs = [math.comb(n, k) for k in range(n + 1)]5print(coeffs)                    # -> [1, 3, 3, 1]6print(sum(coeffs))               # -> 87a, b = 2.0, 1.08print(sum(c * a ** (n - k) * b**k for k, c in enumerate(coeffs)))  # -> 27.0

Coefficients $[1, 3, 3, 1]$ , total 8, value $27.0 = 3^3$ : counting and arithmetic coincide.

Watch Out For

Applying it with a minus and dropping the alternation

$(a - b)^3$ is $(a + (-b))^3$ : odd- $k$ terms go negative, giving $a^3 - 3a^2b + 3ab^2 - b^3$ . Forgetting the alternation and reusing plus signs produces $a^3 + 3a^2b + \cdots$ , wrong wherever $b$ is nonzero.

Substitute $b \to -b$ explicitly before reading coefficients. If an expansion of a difference has all positive terms, the signs were dropped.

Generalizing to non-integer powers term by term

$(a + b)^{1/2}$ is not a four-term polynomial: non-integer exponents need the infinite Newton series with generalized coefficients, which converges only under conditions ( $|b/a| < 1$ for the standard form). Truncating after $n+1$ terms of the finite pattern gives a wrong "exact" answer.

Check the exponent is a non-negative integer before using the finite form. Otherwise reach for the generalized series and state its convergence region.

The Quick Version

$(a+b)^n$ sums $\binom{n}{k} a^{n-k} b^k$ : coefficients count the routes to each mix.
Row $1, 3, 3, 1$ expands the cube: $a^3 + 3a^2b + 3ab^2 + b^3$ .
Coefficients always sum to $2^n$ ; alternating sum is 0.
At $(2, 1)$ the cube gives $27 = 3^3$ , confirmed three ways.
Pascal entries sum from the pair above: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ .
Minuses alternate signs; fractional powers need the infinite series, not this form.