Question

Calcule o valor de n.

Answer 1

Olá Lucas!

Sabemos da definição de Binômio de Newton que:

$\mathsf{(x + y)^n = \sum_{p = 0}^{n} \binom {n}{p} \cdot x^{n - p} \cdot y^p}$

 Isto posto, tiramos que:

$\\ \mathsf{\sum_{p = 0}^{n} \binom {n}{p} = \mathsf{\sum_{p = 0}^{n} \binom {n}{p}} \cdot 1 \cdot 1} \\\\\\ \mathsf{\sum_{p = 0}^{n} \binom {n}{p} = \sum_{p = 0}^{n} \binom {n}{p} \cdot 1^{n - p} \cdot 1^p} \\\\\\ \mathsf{\sum_{p = 0}^{n} \binom {n}{p} = (1 + 1)^n} \\\\\\ \mathsf{\sum_{p = 0}^{n} \binom {n}{p} = 2^n}$

$\\ \mathsf{512 = 2^n} \\\\ \mathsf{2^9 = 2^n} \\\\ \boxed{\boxed{\mathsf{n = 9}}}$

 Boa questão!!