IME / ITA ⇒ (ITA - 2004) Binômio de Newton Tópico resolvido

[Natan]

O termo independente de [tex3]x[/tex3] no desenvolvimento do binômio

• [tex3]\left(\sqrt{\frac{3\sqrt[3]{x}}{5x}}-\sqrt[3]{\frac{5x}{3\sqrt{x}}}\right)^{12}[/tex3]

é:

a) [tex3]729\sqrt[3]{45}[/tex3]
b) [tex3]972\sqrt[3]{15}[/tex3]
c) [tex3]891\sqrt[3]{\frac{3}{5}}[/tex3]
d) [tex3]376\sqrt[3]{\frac{5}{3}}[/tex3]
e) [tex3]165\sqrt[3]{75}[/tex3]

[Karl Weierstrass]

Seja [tex3]T_{k+1}=\left(\begin{array}{cc}n\\ k\end{array}\right)\,\cdot\, a^{n-k}\,\cdot\, b^k[/tex3] o termo geral do binômio [tex3](a+b)^n[/tex3].

Se

• [tex3](a+b)^n=\left(\sqrt{\frac{3\sqrt[3]{x}}{5x}}-\sqrt[3]{\frac{5x}{3\sqrt{x}}}\right)^{12}[/tex3],

então

• [tex3]a=\sqrt{\frac{3x^{\frac{1}{3}-1}}{5}}=
  \left(\frac{3x^
  {-\frac{2}{3}}}{5}\right)^{\frac{1}{2}}=
  \frac{3^{\frac{1}{2}}\,\cdot \,x^{-\frac{1}{3}}}{5^{\frac{1}{2}}}[/tex3],

• [tex3]b=-\sqrt[3]{\frac{5x^{1-\frac{1}{2}}}{3}}=
  -\left(\frac{5x^{\frac{1}{2}}}{3}\right)^{\frac{1}{3}}=
  -\frac{5^{\frac{1}{3}}\,\cdot \,x^{\frac{1}{6}}}{3^{\frac{1}{3}}}[/tex3]

e

• [tex3]n=12[/tex3]

Logo,

• [tex3]T_{k+1}=\left(\begin{array}{cc}12\\ k\end{array}\right)\,\cdot\, \left(\frac{3^{\frac{1}{2}}\,\cdot \,x^{-\,\frac{1}{3}}}{5^{\frac{1}{2}}}\right)^{12-k}\,\cdot\, \left(-\frac{5^{\frac{1}{3}}\,\cdot \,x^{\frac{1}{6}}}{3^{\frac{1}{3}}}\right)^k=
  (-1)^k\,\cdot\,\left(\begin{array}{cc}12\\ k\end{array}\right)\,\cdot\,3^{\frac{36-5k}{6}}\,\cdot\,5^{\frac{5k-36}{6}}\,\cdot\,x^{\frac{k-8}{2}}[/tex3]

Para que exista um termo independente de [tex3]x[/tex3], deve-se ter

• [tex3]\frac{k-8}{2}=0\Longrightarrow k=8.[/tex3]

Portanto,

• [tex3]T_{9}=(-1)^8\,\cdot\,\left(\begin{array}{cc}12\\ 8\end{array}\right)\,\cdot\,3^{-\frac{2}{3}}\,\cdot\,5^{\frac{2}{3}}.[/tex3]

Após algumas manipulações encontramos

• [tex3]T_{9}=165\sqrt[3]{75}[/tex3].