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  "body": "\n## Proof\n\nWe show that the sums satisfy the Fibonacci recurrence, by [[induction|0ASJVEFA9J]] and [[Pascal's rule|0YRG241HQQ]].\n\n1. First, $\\binom{a}{b} = 0$ for natural numbers $b \u003e a$. We prove this by induction on $a$, using the definition of the [[binomial coefficient|092GEF38YD]].\n2. For $a = 0$ and $b \\ge 1$, the definition gives $\\binom{0}{b} = 0$.\n3. Assume the claim for $a$, and let $b \u003e a + 1$. Write $b = c + 1$ with $c \u003e a$. Then $\\binom{a+1}{c+1} = \\binom{a}{c} + \\binom{a}{c+1} = 0 + 0 = 0$.\n4. For a natural number $n$, let $T_n = \\sum_{k=0}^{n} \\binom{n-k}{k}$. If $k \u003e \\lfloor n/2 \\rfloor$, then $2k \u003e n$, so $k \u003e n - k$ and $\\binom{n-k}{k} = 0$ by step 1.\n5. So $T_n = \\sum_{k=0}^{\\lfloor n/2 \\rfloor} \\binom{n-k}{k}$, the right side of the statement.\n6. We have $T_0 = \\binom{0}{0} = 1$ and $T_1 = \\binom{1}{0} + \\binom{0}{1} = 1 + 0 = 1$.\n7. Let $n$ be a natural number. In $T_{n+2}$, the term $k = 0$ is $\\binom{n+2}{0} = 1$. Writing $k = j + 1$ for the other terms gives $T_{n+2} = 1 + \\sum_{j=0}^{n+1} \\binom{n+1-j}{j+1}$.\n8. The term $j = n + 1$ is $\\binom{0}{n+2} = 0$. For $0 \\le j \\le n$, $n - j$ is a natural number, and Pascal's rule gives $\\binom{n+1-j}{j+1} = \\binom{n-j}{j} + \\binom{n-j}{j+1}$.\n9. Hence $T_{n+2} = \\sum_{j=0}^{n} \\binom{n-j}{j} + \\left(1 + \\sum_{j=0}^{n} \\binom{n-j}{j+1}\\right)$.\n10. The first sum is $T_n$. In the bracket, write $i = j + 1$ and $1 = \\binom{n+1}{0}$. The bracket becomes $\\sum_{i=0}^{n+1} \\binom{n+1-i}{i} = T_{n+1}$.\n11. So $T_{n+2} = T_{n+1} + T_n$ for every natural number $n$.\n12. Let $P(n)$ be the claim that $T_n = F_{n+1}$ and $T_{n+1} = F_{n+2}$, where $F_m$ are the [[Fibonacci numbers|0GA9K57A39]].\n13. $P(0)$ holds by step 6, since $F_1 = 1$ and $F_2 = F_1 + F_0 = 1$.\n14. If $P(n)$ holds, then $T_{n+2} = T_{n+1} + T_n = F_{n+2} + F_{n+1} = F_{n+3}$ by step 11. So $P(n + 1)$ holds.\n15. By induction, $F_{n+1} = T_n = \\sum_{k=0}^{\\lfloor n/2 \\rfloor} \\binom{n-k}{k}$ for every natural number $n$.\n\n## Idea\n\nPascal's rule splits each term of the sum for $n + 2$ into one term of the sum for $n + 1$ and one of the sum for $n$. So these diagonal sums of Pascal's triangle satisfy the Fibonacci recurrence. They start with $1, 1$, like $F_1, F_2$.\n",
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