1º) Desenvolva os seguintes binômios de newton:

a) ( x− 2)5

b) ( a+ 3)4

c) (3 − b)4

d) (5 +y )3

Respostas

respondido por: MatiasHP

Olá, siga a explicação abaixo:

Desenvolva os seguintes binômios de newton:

1° Questão:

Sendo:

$(x-2)^{5}$

Adotando a Fórmula de Newton:

$(x+a)^{n} = \sum ^{n}_{p=0} = \left[\begin{array}{c}n\\p\end{array}\right] x^{n-p} a^{p}$

$x= x\\ a=-2\\ n= 5$

Logo:

$(x-2)^{5} \rightarrow \sum ^{5} _{p=0} \left[\begin{array}{cc}5\\p\end{array}\right] x^{5-p} \times -2^{p}$

$\left[\begin{array}{cc}n\\p\end{array}\right] = \frac{n!}{p!(n-p)!}$

Agora para cada, p, obtemos:

P=0

$\frac{n!}{p!(n-p)!} \times x^{(n-p)} \times -2^{p} \rightarrow \frac{5!}{0!(5-0)!} \times x^{(5-0)} \times -2^{0} \rightarrow \frac{5!}{0!5!} \times x^{5} \times -2^{0} \\ \\ \rightarrow \frac{5!}{5!} \times x^{5} \times 1 = x^{5}$

P=1

$\frac{n!}{p!(n-p)!} \times x^{(n-p)} \times -2^{p} \rightarrow \frac{5!}{1!(5-1)!} \times x^{(5-1)} \times -2^{1} \rightarrow \frac{5!}{1!4!} \times x^{4} \times -2^{1} \\ \\ \rightarrow \frac{5\times4}{5!} \times -2x^{4}} \rightarrow 5 \times -2^{4}= -10x^{4}$

P=2

$\frac{n!}{p!(n-p)!} \times x^{(n-p)} \times -2^{p} \rightarrow \frac{5!}{2!(5-2)!} \times x^{(5-2)} \times -2^{2} \rightarrow \frac{5!}{2!3!} \times x^{3} \times 4 \\ \\ Se\:\: mantem: \\ \\ 40x^{3}$

P=3

$\frac{n!}{p!(n-p)!} \times x^{(n-p)} \times -2^{p} \rightarrow \frac{5!}{3!(5-3)!} \times x^{(5-3)} \times -2^{3} \rightarrow \frac{5!}{3!2!} \times x^{2} \times 8 \\ \\ Fica:\\ \\ 80x^{2}$

P=4

$\frac{n!}{p!(n-p)!} \times x^{(n-p)} \times -2^{p} \rightarrow \frac{5!}{4!(5-4)!} \times x^{(5-4)} \times -2^4} \rightarrow \frac{5!}{4!1!} \times x^{1} \times 16 \\ \\ E\:\: igual:\\ \\ 80x^{1}$

P=5

$\frac{n!}{p!(n-p)!} \times x^{(n-p)} \times -2^{p} \rightarrow \frac{5!}{5!(5-5)!} \times x^{(5-5)} \times -2^5} \rightarrow \frac{5!}{5!0!} \times x^{0} \times 32 \\ \\ E\:\: igual:\\ \\ 32$