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  "body": "\n## Statement\n\nLet $r$ be a real number with $r \\ne 1$, and let $n$ be a natural number.\nThen\n$$\\sum_{k=0}^{n-1} r^k = 1 + r + r^2 + \\cdots + r^{n-1} = \\frac{1 - r^n}{1 - r}.$$\n\n## Remarks\n\nThe proof multiplies the sum $S$ by $1 - r$. The product $(1 - r) S$ telescopes to $1 - r^n$.\nFor $r = 1$ the sum is $n$.\nThe identity holds in each commutative ring in the form $(1 - r) \\sum_{k=0}^{n-1} r^k = 1 - r^n$, with no division.\nMathlib also states the version for a sum over the indices $m \\le k \u003c n$, as `geom_sum_Ico`.\nLetting $n \\to \\infty$ gives the [[geometric series|01X5CVV1Q5]].\n\n## Explanations\n\n### L1 human\n\nWrite $S = 1 + r + \\cdots + r^{n-1}$ and $rS = r + r^2 + \\cdots + r^n$.\nSubtract, and almost all terms cancel: $S - rS = 1 - r^n$.\nFor example, $1 + 2 + 4 + 8 = (1 - 16)/(1 - 2) = 15$.\n",
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