Question

Um dos termos de desenvolvimento de (x+1)n , é 12x , determine o valor de n :
${(x + 1)}^{n}$

Answer 1

Olá!

Por Binômio de Newton,

$\\ \displaystyle \mathsf{(x + 1)^n = \sum_{i = 0}^{n} \binom{n}{i} \cdot x^{n - i} \cdot 1^i} \\\\\\ \mathsf{(x + 1)^n = \binom{n}{0} \cdot x^n \cdot 1^0 + \binom{n}{1} \cdot x^{n - 1} \cdot 1^1 + ... + \binom{n}{n - 1} \cdot x^1 \cdot 1^{n - 1} + \binom{n}{n} \cdot x^0 \cdot 1^n} \\\\\\ \mathsf{(x + 1)^n = \binom{n}{0} \cdot x^n + \binom{n}{1} \cdot x^{n - 1} + ... + \underbrace{\mathsf{\binom{n}{n - 1} \cdot x}}_{12x} + \binom{n}{n}}$

Por fim, temos que:

$\\ \displaystyle \mathsf{\binom{n}{n - 1} \cdot x = 12x} \\\\\\ \mathsf{\binom{n}{n - 1} = 12} \\\\ \mathsf{C_{n, n - 1} = 12} \\\\ \mathsf{C_{n, n - 1} = 12} \\\\ \mathsf{\frac{n!}{(n - 1)!\left [ n - \left ( n - 1 \right )! \right ]} = 12} \\\\\\ \mathsf{\frac{n \cdot (n - 1)!}{(n - 1)!1!} = 12} \\\\\\ \boxed{\boxed{\mathsf{n = 12}}}$