Question

reresolva o binomio(x+2)^8

Answer 1

$\left(a+b \right )^{n}=\sum_{k=0}^{n}\dbinom{n}{k}\,a^{n-k}b^{k}$


onde $\dbinom{n}{k}=\dfrac{n!}{k! \cdot \left(n-k \right )!}$ é o coeficiente binomial do $\left(k+1 \right )$-ésimo termo.


Para o binômio $\left(x+2 \right )^{8}$, temos

$a=x\\ \\ b=2\\ \\ n=8\\ \\ \\ \left(x+2 \right )^{8}=\sum_{k=0}^{8}\dbinom{8}{k}\,x^{8-k}\cdot 2^{k}\\ \\ \\ =\dbinom{8}{0}\,x^{8-0}\cdot 2^{0}+\dbinom{8}{1}\,x^{8-1}\cdot 2^{1}+\dbinom{8}{2}\,x^{8-2}\cdot 2^{2}+\dbinom{8}{3}\,x^{8-3}\cdot 2^{3}\\ \\ +\dbinom{8}{4}\,x^{8-4}\cdot 2^{4}+\dbinom{8}{5}\,x^{8-5}\cdot 2^{5}+\dbinom{8}{6}\,x^{8-6}\cdot 2^{6}+\dbinom{8}{7}\,x^{8-7}\cdot 2^{7}\\ \\ +\dbinom{8}{8}\,x^{8-8}\cdot 2^{8}\\ \\ \\ =x^{8}+8x^{7}\cdot 2+28x^{6} \cdot 2^{2}+56x^{5}\cdot 2^{3}\\ +70x^{4}\cdot 2^{4}+56x^{3}\cdot 2^{5}+28x^{2}\cdot 2^{6}+8x\cdot 2^{7}+2^{8}\\ \\ \\ =x^{8}+16x^{7}+112x^{6}+448x^{5}+1\,120x^{4}+1\,792x^{3}+1\,792x^{2}+1\,024x+256$