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  "body": "\n## Proof\n\nWe argue by [[induction|0ASJVEFA9J]] on $n$.\nFor each natural number $n$, let $T_n = \\sum_{k=0}^{n} k$, and let $P(n)$ be the claim $T_n = \\frac{n(n+1)}{2}$.\nWe compute with rational numbers, so division by $2$ is allowed.\n\n1. Base case: $T_0 = 0$, and $\\frac{0 \\cdot 1}{2} = 0$. So $P(0)$ holds.\n2. Induction step: suppose that $P(n)$ holds for some natural number $n$.\n3. The sum up to $n + 1$ is the sum up to $n$ plus the last term: $T_{n+1} = T_n + (n + 1)$.\n4. By the hypothesis, $T_{n+1} = \\frac{n(n+1)}{2} + (n + 1)$.\n5. We rewrite: $\\frac{n(n+1)}{2} + \\frac{2(n+1)}{2} = \\frac{(n+1)(n+2)}{2}$, by factoring out $n + 1$.\n6. So $T_{n+1} = \\frac{(n+1)((n+1)+1)}{2}$, which is $P(n + 1)$.\n7. By induction, $P(n)$ holds for every natural number $n$, that is, $0 + 1 + \\dots + n = \\frac{n(n+1)}{2}$.\n\n## Idea\n\nGoing from $n$ to $n + 1$ adds the term $n + 1$ to the sum.\nThe formula grows by the same amount, because $\\frac{(n+1)(n+2)}{2} - \\frac{n(n+1)}{2} = n + 1$.\n",
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