Question

No desenvolvimento de
${(2x - \frac{1}{x}) }^{22}$
determine:

a) o coeficiente do termo em x¹⁸
b) o termo independente de x

respostas: a: 231 x 10²⁰

b:
$- \binom{22}{11} \times {2}^{11}$
​

Answer 1

Olá Gustavo!

Resposta:

$\\ \displaystyle \bullet \qquad \boxed{\mathtt{231 \cdot 2^{20}}} \\\\ \bullet \qquad \boxed{\mathtt{- 2^{11} \cdot \binom{22}{11}}}$

Explicação passo-a-passo:

Dos conceitos envolvendo o Binômio de Newton, temos que:

$\displaystyle \mathtt{(x + y)^n = \binom{n}{0} x^{n} \cdot y^{0} + \binom{n}{1} x^{n - 1} \cdot y^{1} + ... + \binom{n}{p - 1} x^{n - p + 1} \cdot y^{p - 1} + \binom{n}{p} x^{n - p} \cdot y^{p}}$

a) seja k um inteiro positivo, tal que, no desenvolvimento do binômio, ele retorna o coeficiente do termo de $\displaystyle \mathtt{x^{18}}$. Veja:

$\displaystyle \mathtt{\left ( 2x - \frac{1}{x} \right )^{22} = \binom{22}{0} (2x)^{22} \cdot \left ( - \frac{1}{x} \right )^0 + ... + \binom{22}{k} (2x)^{k - 22} \cdot \left ( - \frac{1}{x} \right )^k + ... + \binom{22}{22} (2x)^{0} \cdot \left ( - \frac{1}{x} \right )^{22}}$

Desenvolvendo o termo,

$\\ \displaystyle \mathsf{\binom{22}{k} (2x)^{22 - k} \cdot \left ( - \frac{1}{x} \right )^k = \binom{22}{k} 2^{22 - k} \cdot x^{22 - k} \cdot \left ( - x^{- 1} \right )^k} \\\\\\ \mathsf{\qquad \qquad \qquad \qquad \qquad \quad = \binom{22}{k} 2^{22 - k} \cdot x^{22} \cdot x^{- k} \cdot (- x)^{- k}} \\\\\\ \mathsf{\qquad \qquad \qquad \qquad \qquad \quad = \binom{22}{k} 2^{22 - k} \cdot x^{22} \cdot \left [ x \cdot (- x) \right ]^{- 2k}} \\\\\\ \mathsf{\qquad \qquad \qquad \qquad \qquad \quad = \binom{22}{k} 2^{22 - k} \cdot x^{22} \cdot (- x)^{- 2k}}$

Ora, o expoente de x deve ser 18, então, podemos afirmar que o expoente de - x é par! Daí,

$\\ \displaystyle \mathsf{\binom{22}{k} (2x)^{22 - k} \cdot \left ( - \frac{1}{x} \right )^k = \binom{22}{k} 2^{22 - k} \cdot x^{22} \cdot (- x)^{- 2k}} \\\\\\ \mathsf{\qquad \qquad \qquad \qquad \qquad \ = \binom{22}{k} 2^{22 - k} \cdot x^{22} \cdot x^{- 2k}} \\\\\\ \mathsf{\qquad \qquad \qquad \qquad \qquad \ = \binom{22}{k} 2^{22 - k} \cdot x^{22 - 2k}}$

Com efeito,

$\\ \displaystyle \mathsf{22 - 2k = 18} \\ \mathsf{22 - 18 = 2k} \\ \mathsf{4 = 2k} \\ \boxed{\mathsf{k = 2}}$

Portanto,

$\\ \displaystyle \mathsf{\binom{22}{k} 2^{22 - k} \cdot x^{22 - 2k} \xrightarrow{k = 2} \binom{22}{2} 2^{22 - 2} \cdot x^{22 - 2 \cdot 2}} \\\\\\ \mathsf{\qquad \ \qquad \ \qquad \xrightarrow{k = 2}\frac{22!}{(22 - 2)!2!} \cdot 2^{20} \cdot x^{22 - 4}} \\\\\\ \mathsf{\qquad \ \qquad \xrightarrow{k = 2} \frac{22 \cdot 21 \cdot 20!}{20!2 \cdot 1} \cdot 2^{20} \cdot x^{18}} \\\\\\ \mathsf{\qquad \xrightarrow{k = 2} \left ( 11 \cdot 21 \right ) \cdot 2^{20} \cdot x^{18}} \\\\\\ \mathsf{\mathsf{\xrightarrow{k = 2} \boxed{\boxed{\mathsf{231 \cdot 2^{20} \cdot x^{18}}}}}}$

b) Lembrando que o termo independente é aquele não possui variável/incógnita. Noutras palavras, quando o expoente de x for nulo! Isto posto, em

$\displaystyle \mathsf{\binom{22}{k} (2x)^{22 - k} \cdot \left ( - \frac{1}{x} \right )^k = \binom{22}{k} 2^{22 - k} \cdot x^{22 - k} \cdot (- x)^{- k}}$

Fazemos,

$\\ \displaystyle \mathsf{22 - k - k = 0} \\ \mathsf{22 = 2k} \\ \boxed{\mathsf{k = 11}}$

Logo,

$\\ \displaystyle \mathsf{\binom{22}{k} 2^{22 - k} \cdot x^{22 - k} \cdot (- x)^{- k} \xrightarrow{k = 11} \binom{22}{11} 2^{22 - 11} \cdot x^{22 - 11} \cdot (- x)^{- 11}} \\\\\\ \mathsf{\qquad \ \qquad \ \qquad \xrightarrow{k = 11} \binom{22}{11} 2^{11} \cdot x^{11} \cdot (- x)^{- 11}} \\\\\\ \mathsf{\qquad \ \qquad \xrightarrow{k = 11} \binom{22}{11} 2^{11} \cdot (- 1) \cdot x^0} \\\\\\ \mathsf{\mathsf{\xrightarrow{k = 11} \boxed{\boxed{- 2^{11} \cdot \binom{22}{11}}}}}$