Use a Fórmula (17) da série binomial para obter a série de Maclaurin de
\sqrt[3]{1+x}

Passo 1

O enunciado pede a série de Maclaurin de
f(x)=\sqrt[3]{1+x}=(1+x)^{\frac13}
Usaremos a fórmula da série binomial generalizada (Fórmula (17)):
(1+u)^\alpha=\sum_{n=0}^{\infty}\binom{\alpha}{n}u^n,
onde
\binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}
e a expansão converge para \lvert u\rvert<1.

Passo 2

Identificamos na nossa função \alpha=\tfrac13 e u=x. Logo
(1+x)^{\frac13}=\sum_{n=0}^{\infty}\binom{\tfrac13}{n}x^n,
com
\binom{\tfrac13}{n} =\frac{\tfrac13\bigl(\tfrac13-1\bigr)\bigl(\tfrac13-2\bigr)\cdots\bigl(\tfrac13-n+1\bigr)}{n!}.

Passo 3

Vamos calcular os primeiros coeficientes \binom{\tfrac13}{n}:
n =0:
\binom{\tfrac13}{0}=1.
n =1:
\binom{\tfrac13}{1}=\frac{\tfrac13}{1!}=\frac13.
n =2:
\binom{\tfrac13}{2} =\frac{\tfrac13\bigl(\tfrac13-1\bigr)}{2!} =\frac{\tfrac13\cdot(-\tfrac23)}{2} =-\frac{2}{9}.
n =3:
\binom{\tfrac13}{3} =\frac{\tfrac13\cdot(-\tfrac23)\cdot(-\tfrac53)}{3!} =\frac{1\cdot(-2)\cdot(-5)}{3^3\cdot6} =\frac{10}{162} =\frac{5}{81}.
n =4:
\binom{\tfrac13}{4} =\frac{\tfrac13\cdot(-\tfrac23)\cdot(-\tfrac53)\cdot(-\tfrac83)}{4!} =-\frac{1\cdot2\cdot5\cdot8}{3^4\cdot24} =-\frac{80}{1944} =-\frac{10}{243}.

Passo 4

Substituindo esses coeficientes na soma infinita, obtemos a série de Maclaurin:
\sqrt[3]{1+x} =1+\frac13x-\frac{2}{9}x^2+\frac{5}{81}x^3-\frac{10}{243}x^4+\cdots.

Resposta

\sqrt[3]{1+x} =1+\frac13x-\frac{2}{9}x^2+\frac{5}{81}x^3-\frac{10}{243}x^4+\cdots.

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