CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26

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$(n + 1) C_{n} = (\binom{2 n}{n})$

Statement

For every $n \in N$ , in $N$ ,

$(n + 1) C_{n} = (\binom{2 n}{n}) .$

Facts & Assumptions

Given: a natural number $n$ .

[F1]

$C_{n} + (\binom{2 n}{n + 1}) = (\binom{2 n}{n})$ in $N$ ( $C_{n} + (\binom{2 n}{n + 1}) = (\binom{2 n}{n})$ ).

[L1]

For $m, k \in N$ with $k \leq m$ : $(\binom{m}{k}) \cdot k! \cdot (m - k)! = m!$ in $N$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

[L2]

$m! \neq 0$ for every $m \in N$ , and $σ (m)! = m! \cdot σ (m)$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L3]

$(\binom{m}{k}) = 0$ for $k > m$ , and $(\binom{m}{0}) = 1$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L4]

For all $x, y, k \in N$ with $k \neq 0$ : if $x \cdot k = y \cdot k$ then $x = y$ (Cancellation for multiplication by a nonzero factor).

[L5]

For all $x, y, k \in N$ : if $x + k = y + k$ then $x = y$ (Addition is cancellative).

Proof

technique · direct

1.1L1L2L3L4algebra

First, $(n + 1) (\binom{2 n}{n + 1}) = n (\binom{2 n}{n})$ in $N$ . For $n = 0$ both sides are $0$ , since $(\binom{0}{1}) = 0$ by [L3]. For $n \geq 1$ one has $n + 1 \leq 2 n$ and $2 n - (n + 1) = n - 1$ , so [L1] gives $(\binom{2 n}{n + 1}) (n + 1)! (n - 1)! = (2 n)!$ and $(\binom{2 n}{n}) n! n! = (2 n)!$ ; writing $(n + 1)! = (n + 1) \cdot n!$ and the second factor $n! = n \cdot (n - 1)!$ by [L2], the two left sides read $(n + 1) (\binom{2 n}{n + 1}) \cdot n! (n - 1)!$ and $n (\binom{2 n}{n}) \cdot n! (n - 1)!$ , and cancelling the nonzero factor $n! (n - 1)!$ by [L2] and [L4] gives the identity.

2.1F1L3L5step 1.1algebra∎

Multiply [F1] by $n + 1$ : $(n + 1) C_{n} + (n + 1) (\binom{2 n}{n + 1}) = (n + 1) (\binom{2 n}{n})$ . By step 1.1 the second summand on the left is $n (\binom{2 n}{n})$ , and the right-hand side is $n (\binom{2 n}{n}) + (\binom{2 n}{n})$ , so cancelling the common summand $n (\binom{2 n}{n})$ by [L5] gives $(n + 1) C_{n} = (\binom{2 n}{n})$ . At $n = 0$ this reads $C_{0} = (\binom{0}{0}) = 1$ .

Remarks

The quotient form. The identity is an identity of natural numbers. Reading each natural number as its canonical natural in $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field) and dividing by the nonzero real $n + 1$ turns it into the familiar $C_{n} = \frac{1}{n + 1} (\binom{2 n}{n})$ . The multiplicative form is the one proved, and it is the form in which no division and no embedding is needed; it also says at once that $n + 1$ divides the central binomial coefficient, which the quotient form presupposes.
What the proof actually uses. Only the reflection identity and factorial bookkeeping. The Catalan number is never manipulated as a formula: it enters as the count it was defined to be and leaves as a factor of a binomial coefficient.

Depends on

Used by

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. Guichard, An Introduction to Combinatorics and Graph Theory, §3.5 Catalan Numbers (standard reference, not scraped)
C. Krattenthaler, "Lattice Path Enumeration", ch. 10 of the Handbook of Enumerative Combinatorics, §10.3 (standard reference, not scraped)

(n+1) Cn=(2nn)(n+1)Cn​=(n2n​)