Pré-Vestibular ⇒ (AFA 2006) Binômio de Newton

[Natan]

Os 3 primeiros coeficientes no desenvolvimento de [tex3]\left(x^2+\frac{1}{2x}\right)^{n}[/tex3] estão em P.A. O valor de [tex3]n[/tex3] é:

[tex3]a)\, 4 \\ b)\, 6 \\ c)\, 8 \\ d)\, 10 \\ e)\, 12[/tex3]

[joynobre]

[tex3](x+a)^n \Rightarrow\,\,T_{p+1}=\(n\\p\).x^{n-p}.a^p[/tex3]

[tex3]\left(x^2+\frac{1}{2x}\right)^{n}[/tex3]

[tex3]T_{p+1}=\(n\\p\).x^{2(n-p)}.(2x)^{-1p}[/tex3]
[tex3]T_{p+1}=\(n\\p\).x^{2n-3p}.2^{-p}[/tex3]

[tex3]T_1=\(n\\0\)x^{2n}.2^0[/tex3]
[tex3]T_2=\(n\\1\)x^{2n-3}.2^{-1}[/tex3]
[tex3]T_3= \(n\\2\)x^{2n-6}. 2^{-2}[/tex3]

P.A.[tex3]\left( \(n\\0\),\,\, \(n\\1\).2^{-1},\,\, \(n\\2\).2^{-2} \right)\Rightarrow \,\, \left(1,\,\,\frac{n}{2},\,\, \frac{n(n-1)}{8}\right)[/tex3]

[tex3]\frac{n}{2}=\frac{1+\frac{n(n-1)}{8}}{2}[/tex3]
[tex3]\frac{n}{2}=\frac{n^2-n+8}{16}[/tex3]
[tex3]n^2-9n+8=0[/tex3]
[tex3]n=8 ou[/tex3] [tex3]n=1[/tex3] (não convém)

[tex3]\therefore \,\, n=8[/tex3]