Question

Questão.5 Calcule o valor da spressão
63 + 3) usando o desenvolvimento binomial de
Newton.

Answer 1

$(a+b)^n=\binom{n}{0}.a^n.b^0+\binom{n}{1}.a^{n-1}.b^1+...+\binom{n}{n}.a^0.b^n\\\\ (\sqrt{3}+3)^5=\binom{5}{0}.(\sqrt{3})^5.3^0+\binom{5}{1}.(\sqrt{3})^4.3^1+\binom{5}{2}.(\sqrt{3})^3.3^2+\\ +\binom{5}{3}.(\sqrt{3})^2.3^3+\binom{5}{4}.(\sqrt{3})^1.3^4+\binom{5}{5}.(\sqrt{3})^0.3^5$

$\binom{5}{0}=\frac{5!}{0!(5-0)!}=\frac{5!}{5!}=1\ \| \ \binom{5}{1}=\frac{5!}{1!(5-1)!}=\frac{5.4!}{4!}=5\\ \binom{5}{2}=\frac{5!}{2!(5-2)!}=\frac{5.4.3!}{2.3!}=10\ \| \binom{5}{3}=\frac{5!}{3!(5-3)!}=\frac{5.4.3!}{3!.2}=10\\ \binom{5}{4}=\frac{5!}{4!(5-4)!}=\frac{5.4!}{4!}=5\ \| \binom{5}{5}=\frac{5!}{5!(5-5)!}=\frac{5!}{5!}=1\\\\ (\sqrt{3}+3)^5=(\sqrt{3})^5+5.(\sqrt{3})^4.3+10.(\sqrt{3})^3.9\\ +10.(\sqrt{3})^2.27+5.(\sqrt{3})^1.81+243\\\\ (\sqrt{3}+3)^5=9\sqrt{3}+135+270\sqrt{3}+810+405\sqrt{3}+243$

$\| \ (\sqrt{3}+3)^5=684\sqrt{3}+1188\ \|$