Binomial Theorem

(a+b)^n = _k nk a^n-k b^k — coefficients come from choices.

The idea

Theorem (Binomial theorem).

For any numbers $a$ and $b$ and any whole number $n \ge 0$, $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}\, a^{n-k} b^{k}.$

The figure shows the case $n = 2$ as areas: a square of side $x + a$ splits into an $x^{2}$ block, an $a^{2}$ block, and two congruent $ax$ rectangles — the coefficient $\binom{2}{1} = 2$, counted as the two ways one factor can supply the $a$.

Each coefficient is a count. Write the power as $n$ separate factors, $(a+b)(a+b)\cdots(a+b)$. Expanding by the distributive law means picking one letter, $a$ or $b$, from each factor, multiplying the $n$ picks together, and adding the results over all $2^n$ ways to pick.

The objects being counted are these picks, and two picks produce the same term exactly when they take $b$ from the same number of factors: multiplication does not record which factors supplied the $b s, so every pick taking $b$ from exactly $k$ factors collapses to the single monomial $a^{n-k}b^{k}$. The number of such picks is the number of ways to choose $k$ of the $n$ factors, namely $\binom{n}{k}$. Collecting equal monomials leaves each with its count in front, which is the formula.

Substituting particular values for $a$ and $b$ turns the one theorem into many identities.

Ways to work on it

• Walkthrough. Why the formula has nk in it, plus a full expansion term by term.
• Proof. See why — the combinatorial argument: each term of (x+y)^n counts a choice of factors.
• Practice. Extract a single coefficient from (ax + b)^n.
• Hardest. Match an identity to the right substitution into (a+b)^n.

Not sure where to start? Take the ten-question placement test.