Question

alguém responde isso por favor, é urgente... sobre binômios de Newton...

Answer 1

Binômio de Newton:

$\boxed{\boxed{(a+b)^{n}=\sum\limits_{k=0}^{n}C_{n,k}\cdot a^{n-k}\cdot b^{k}}}$

onde

$C_{n,k}=\dfrac{n!}{k!\cdot(n-k)!}$
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Calculando as combinações do binômio anteriormente:

$C_{4,0}=\dfrac{4!}{0!(4-0)!}=\dfrac{4!}{4!}=1\\\\\\C_{4,1}=\dfrac{4!}{1!(4-1)!}=\dfrac{4!}{3!}=\dfrac{4\cdot3!}{3!}=4\\\\\\C_{4,2}=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!2!}=\dfrac{4\cdot3\cdot2!}{2\cdot1\cdot2!}=6\\\\\\C_{4,3}=\dfrac{4!}{3!(4-3)!}=\dfrac{4!}{3!1!}=\dfrac{4\cdot3!}{3!}=4\\\\\\C_{4,4}=\dfrac{4!}{4!(4-4)!}=\dfrac{4!}{4!0!}=\dfrac{4!}{4!}=1$
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$(y+3)^{4}$

Comparando com a fórmula, temos que

a = y
b = 3
n = 4

Logo:

$\displaystyle(y+3)^{4}=\sum_{k=0}^{4}C_{4,k}\cdot y^{4-k}\cdot3^{k}=u_{0}+u_{1}+u_{2}+u_{3}+u_{4}$

Vou desenvolver os termos separadamente para dar espaço:

$u_{0}=C_{4,0}\cdot y^{4-0}\cdot3^{0}=1\cdot y^{4}\cdot1=y^{4}\\\\u_{1}=C_{4,1}\cdot y^{4-1}\cdot3^{1}=4\cdot y^{3}\cdot3=12y^{3}\\\\u_{2}=C_{4,2}\cdot y^{4-2}\cdot3^{2}=6\cdot y^{2}\cdot9=54y^{2}\\\\u_{3}=C_{4,3}\cdot y^{4-3}\cdot3^{3}=4\cdot y^{1}\cdot27=108y\\\\u_{4}=C_{4,4}\cdot y^{4-4}\cdot3^{4}=1\cdot y^{0}\cdot81=81$

Portanto:

$\boxed{\boxed{(y+3)^{4}=y^{4}+12y^{3}+54y^{2}+108y+81}}$