Binomial theorem

Statement

Let $R$ be a commutative ring, let $x, y \in R$, and let $n$ be a natural number. Then $$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}.$$ Here $\binom{n}{k}$ is the binomial coefficient, and the integer $\binom{n}{k}$ acts on $R$ by repeated addition.

Remarks

For $n = 2$ it gives $(x + y)^2 = x^2 + 2xy + y^2$. For $n = 3$ it gives $x^3 + 3x^2 y + 3x y^2 + y^3$. The formula stays true in a noncommutative ring when $xy = yx$. With $x = y = 1$ it gives the sum of binomial coefficients. With $x = -1$ and $y = 1$ it gives the alternating sum. The multinomial theorem extends it to sums of more than two terms.

Explanations

L1 human

When you expand $(x + y)^n$, each term comes from choosing $x$ or $y$ in each of the $n$ factors. The term $x^k y^{n-k}$ appears once for each choice of the $k$ factors that give $x$, so its coefficient is $\binom{n}{k}$.