Question

(1+3a/2) elevado a 6 "-".como resolve?

Answer 1

Pelo Binômio de Newton:

$(a+b)^{n}=\dbinom{n}{0}a^n+\dbinom{n}{1}a^{n-1}b^1+\dots+\dbinom{n}{n-1}a^1b^{n-1}+\dbinom{n}{n}b^n$.

Deste modo:

$\left(1+\dfrac{3a}{2}\right)^6=\dbinom{6}{0}1^6+\dbinom{6}{1}1^5\cdot\left(\dfrac{3a}{2}\right)^1+\dbinom{6}{2}1^4\cdot\left(\dfrac{3a}{2}\right)^2+\dbinom{6}{3}1^3\cdot\left(\dfrac{3a}{2}\right)^3+\dbinom{6}{4}1^2\cdot\left(\dfrac{3a}{2}\right)^4+\dbinom{6}{5}1^1\cdot\left(\dfrac{3a}{2}\right)^5+\dbinom{6}{6}\left(\dfrac{3a}{2}\right)^6$.

São esses os sete termos:

$\dbinom{6}{0}1^6=1$

$\dbinom{6}{1}1^5\cdot\left(\dfrac{3a}{2}\right)^1=6\cdot\left(\dfrac{3a}{2}\right)=9a$

$\dbinom{6}{2}1^4\cdot\left(\dfrac{3a}{2}\right)^2=15\cdot\left(\dfrac{9a^2}{4}\right)=\dfrac{135a^2}{4}$

$\dbinom{6}{3}1^3\cdot\left(\dfrac{3a}{2}\right)^3=20\cdot\left(\dfrac{27a^3}{8}\right)=\dfrac{135a^3}{2}$

$\dbinom{6}{4}1^2\cdot\left(\dfrac{3a}{2}\right)^4=15\cdot\left(\dfrac{81a^4}{16}=\dfrac{1215a^4}{16}$

$\dbinom{6}{5}1^1\cdot\left(\dfrac{3a}{2}\right)^5=6\cdot\left(\dfrac{243a^5}{32}\right)=\dfrac{729a^5}{16}$

$\dbinom{6}{6}\cdot\left(\dfrac{3a}{2}\right)^6=\dfrac{729a^6}{64}$.


Portanto,

$\left(1+\dfrac{3a}{2}\right)^6=1+9a+\dfrac{135a^2}{4}+\dfrac{135a^3}{2}+\dfrac{1215a^4}{16}+\dfrac{729a^5}{16}+\dfrac{729a^6}{64}$.