English
For all n,k: (n+1) * C(n,k) = (k+1) * C(n+1, k+1).
Русский
Для любых n,k: (n+1) \\, C(n,k) = (k+1) \\, C(n+1, k+1).
LaTeX
$$$\\\\binom{n}{k} (k+1) = \\\\binom{n+1}{k+1} (k+1)$$$
Lean4
theorem succ_mul_choose_eq : ∀ n k, succ n * choose n k = choose (succ n) (succ k) * succ k
| 0, 0 => by decide
| 0, k + 1 => by simp [choose]
| n + 1, 0 => by simp [choose, mul_succ, Nat.add_comm]
| n + 1, k + 1 => by
rw [choose_succ_succ (succ n) (succ k), Nat.add_mul, ← succ_mul_choose_eq n, mul_succ, ← succ_mul_choose_eq n,
Nat.add_right_comm, ← Nat.mul_add, ← choose_succ_succ, ← succ_mul]