ThmDex – An index of mathematical definitions, results, and conjectures.
Relationship with the sum of initial natural numbers and the binomial coefficient
Formulation 0
Let $N \in \mathbb{N}$ be a D996: Natural number.
Then \begin{equation} \sum_{n = 0}^N n = \binom{N + 1}{2} \end{equation}
Proofs
Proof 0
Let $N \in \mathbb{N}$ be a D996: Natural number.
Result R2782: Arithmetic expression for sum of initial natural numbers shows that \begin{equation} \sum_{n = 0}^N n = \frac{N (N + 1)}{2} \end{equation} while from result R1831: Real arithmetic expression for binomial coefficient, we have \begin{equation} \binom{N + 1}{2} = \frac{(N + 1) !}{(N + 1 - 2) ! 2 !} = \frac{(N + 1) !}{(N - 1) ! 2 !} = \frac{N (N + 1)}{2} \end{equation} $\square$