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  "body": "\n## Statement\n\nLet $R$ be a commutative ring, let $x, y \\in R$, and let $n$ be a natural number.\nThen\n$$(x + y)^n = \\sum_{k=0}^{n} \\binom{n}{k} x^k y^{n-k}.$$\nHere $\\binom{n}{k}$ is the [[binomial coefficient|092GEF38YD]], and the integer $\\binom{n}{k}$ acts on $R$ by repeated addition.\n\n## Remarks\n\nFor $n = 2$ it gives $(x + y)^2 = x^2 + 2xy + y^2$. For $n = 3$ it gives $x^3 + 3x^2 y + 3x y^2 + y^3$.\nThe formula stays true in a noncommutative ring when $xy = yx$.\nWith $x = y = 1$ it gives the [[sum of binomial coefficients|0KAMF19NV8]]. With $x = -1$ and $y = 1$ it gives the [[alternating sum|0PQQGFAWSH]].\nThe [[multinomial theorem|0XFC1VQYEK]] extends it to sums of more than two terms.\n\n## Explanations\n\n### L1 human\n\nWhen you expand $(x + y)^n$, each term comes from choosing $x$ or $y$ in each of the $n$ factors.\nThe term $x^k y^{n-k}$ appears once for each choice of the $k$ factors that give $x$, so its coefficient is $\\binom{n}{k}$.\n",
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