Question

desenvolva o binomio de newton: (x+3)4

Answer 1

Formula do desenvolvimento do Binômio de Newton:

$\displaystyle\mathtt{(a+b)^n=\sum_{p=0}^n\binom{n}{p}\cdot a^{n-p}\cdot b^p}$


sendo $\mathtt{\dbinom{n}{p}=\dfrac{n!}{p!\cdot (n-p)!}}$ o coeficiente binomial.

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O termo da posição $\mathtt{(p+1)}$ é dado por

$\mathtt{t_{p+1}=\dbinom{n}{p}\cdot a^{n-p}\cdot b^p}.$

sendo $\mathtt{0\le p\le n},$
 

de forma que a expansão do binômio de Newton nada mais é do que a soma de todos os termos:

$\displaystyle\mathtt{(a+b)^n=t_1+t_2+\ldots+t_n+t_{n+1}}\\\\ \mathtt{(a+b)^n=\sum_{p=0}^n t_{p+1}}$

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Para o binômio em questão, temos

$\mathtt{(x+3)^4}$

$\mathtt{a=x,\;b=3,\;n=4.}$


•    Calculando o 1º termo:

$\mathtt{t_1=\dbinom{4}{0}\cdot x^{4-0}\cdot 3^0}\\\\\\ \mathtt{t_1=\dfrac{4!}{0!\cdot (4-0)!}\cdot x^4\cdot 3^0}\\\\\\ \mathtt{t_1=\dfrac{\diagup\!\!\!\!\! 4!}{1\cdot \diagup\!\!\!\!\! 4!}\cdot x^4\cdot 1}\\\\\\ \mathtt{t_1=1\cdot x^4\cdot 1}$

$\mathtt{t_1=x^4}$


•    Calculando o 2º termo:

$\mathtt{t_2=\dbinom{4}{1}\cdot x^{4-1}\cdot 3^1}\\\\\\ \mathtt{t_2=\dfrac{4!}{1!\cdot (4-1)!}\cdot x^3\cdot 3^1}\\\\\\ \mathtt{t_2=\dfrac{4\cdot \diagup\!\!\!\!\! 3!}{1!\cdot \diagup\!\!\!\!\! 3!}\cdot x^3\cdot 3}\\\\\\ \mathtt{t_2=4\cdot x^3\cdot 3}$

$\mathtt{t_2=12x^3}$


•    Calculando o 3º termo:

$\mathtt{t_3=\dbinom{4}{2}\cdot x^{4-2}\cdot 3^2}\\\\\\ \mathtt{t_3=\dfrac{4!}{2!\cdot (4-2)!}\cdot x^2\cdot 3^2}\\\\\\ \mathtt{t_3=\dfrac{4\cdot 3\cdot \diagup\!\!\!\!\!2!}{2!\cdot \diagup\!\!\!\!\! 2!}\cdot x^2\cdot 9}\\\\\\ \mathtt{t_3=6\cdot x^2\cdot 9}$

$\mathtt{t_3=54x^2}$


•    Calculando o 4º termo:

$\mathtt{t_4=\dbinom{4}{3}\cdot x^{4-3}\cdot 3^3}\\\\\\ \mathtt{t_4=\dfrac{4!}{3!\cdot (4-3)!}\cdot x^1\cdot 3^3}\\\\\\ \mathtt{t_4=\dfrac{4\cdot \diagup\!\!\!\!\! 3!}{\diagup\!\!\!\!\! 3!\cdot 1!}\cdot x^1\cdot 3^3}\\\\\\ \mathtt{t_4=4\cdot x\cdot 27}$

$\mathtt{t_4=108x}$


•    Calculando o 5º termo:

$\mathtt{t_5=\dbinom{4}{4}\cdot x^{4-4}\cdot 3^4}\\\\\\ \mathtt{t_5=\dfrac{4!}{4!\cdot (4-4)!}\cdot x^0\cdot 3^4}\\\\\\ \mathtt{t_5=\dfrac{\diagup\!\!\!\!\! 4!}{\diagup\!\!\!\!\! 4!\cdot 0!}\cdot x^0\cdot 3^4}\\\\\\ \mathtt{t_5=1\cdot 1\cdot 81}$

$\mathtt{t_5=81}$

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Então, o desenvolvimento do binômio pedido é

$\mathtt{(x+3)^4=t_1+t_2+t_3+t_4+t_5}\\\\ \boxed{\begin{array}{c}\mathtt{(x+3)^4=x^4+12x^3+54x^2+108x+81} \end{array}}$


Dúvidas? Comente.


Bons estudos! :-)