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Fungrim entry: 6e1f13

(z+1k+1)=z+1k+1(zk){z + 1 \choose k + 1} = \frac{z + 1}{k + 1} {z \choose k}
Assumptions:zC  and  kZ0z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; k \in \mathbb{Z}_{\ge 0}
{z + 1 \choose k + 1} = \frac{z + 1}{k + 1} {z \choose k}

z \in \mathbb{C} \;\mathbin{\operatorname{and}}\; k \in \mathbb{Z}_{\ge 0}
Fungrim symbol Notation Short description
Binomial(nk){n \choose k} Binomial coefficient
CCC\mathbb{C} Complex numbers
ZZGreaterEqualZn\mathbb{Z}_{\ge n} Integers greater than or equal to n
Source code for this entry:
    Formula(Equal(Binomial(Add(z, 1), Add(k, 1)), Mul(Div(Add(z, 1), Add(k, 1)), Binomial(z, k)))),
    Variables(z, k),
    Assumptions(And(Element(z, CC), Element(k, ZZGreaterEqual(0)))))

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Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2021-03-15 19:12:00.328586 UTC