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  "body": "\n## Statement\n\nLet $x$ be a real number with $x \\neq 1$, and let $n$ be a natural number.\nThen\n$$\\sum_{k=0}^{n-1} x^k = 1 + x + x^2 + \\dots + x^{n-1} = \\frac{x^n - 1}{x - 1}.$$\n\n## Remarks\n\nThe same formula holds for $x$ in any field, for example for complex numbers, when $x \\neq 1$.\nFor $x = 1$, the sum is $n$.\nA short proof: $(x - 1)(1 + x + \\dots + x^{n-1}) = x^n - 1$, because the other terms cancel.\n\n## Explanations\n\n### L1 human\n\nIn a geometric progression, you get each term from the one before by multiplying by the same number $x$.\nFor $x = 2$, the formula gives $1 + 2 + 4 + 8 = 2^4 - 1 = 15$.\nMultiply the sum by $x$ and subtract the sum: almost all terms cancel, and only $x^n - 1$ stays.\n",
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