Concursos Públicos ⇒ SEQUÊNCIA-QUESTÃO DE CONCURSO

[luciopiupiu]

Resposta

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A

Seja a sequência formada pelos seguintes termos
a_n=(1/2 ,3/2^2 ,5/2^3 ,… ,(2n-1)/2^n ), em que n ∈ {1, 2, 3, ..., n}. Assinale a
alternativa que indique corretamente a soma dos termos de a_n.
A) 3-(2n+3)/2^n

B) 3/2-(2n+3)/2^n

C) 3-(2n+3)/2^(n-1)

D) 1-(2n+2)/2^n

E) 3+(2n-3)/2^n

[PeterPark]

[tex3]\bf S(soma)~~DO~~ENUNCIADO: \\S = \frac{1}{2}+ \frac{3}{2^2}+ \frac{5}{2^3}+\frac{7}{2^4}\dots +\frac{2n-3}{2^{n-1}}+\frac{2n-1}{2^n} ~~~~~~*eq_1\\\ \\ \\\ \\ \\
MULTIPLICANDO ~~S~~ POR~~\frac{1}{2}: \\S\cdot \frac{1}{2} = \frac{1}{2^2}+ \frac{3}{2^3}+ \frac{5}{2^4}\dots +\frac{2n-3}{2^{n}}+\frac{2n-1}{2^{n+1}}~~~~~~*eq_2 \\\ \\ \\\ \\ \\ SUBTRAINDO~~AS~~DUAS~~EQUAÇÕES(eq_1-eq_2): \\
S-S\cdot\frac{1}{2} = \frac{1}{2}+\frac{3}{2^2}-\frac{1}{2^2}+\frac{5}{2^3}-\frac{3}{2^3}+\frac{7}{2^4}-\frac{5}{2^4} \dots +\frac{2n-1}{2^n}-\frac{2n-3}{2^n}-\(\frac{2n-1}{2^{n+1}}\) = \\
=\frac{1}{2}+\frac{1}{2^2}\cdot(3-1)+\frac{1}{2^3}\cdot(5-3)+\frac{1}{2^4}\cdot(7-5)+\dots+\frac{1}{2^n}\cdot(2n-1-2n+3)-\(\frac{2n-1}{2^{n+1}}\) = \\
=\frac{1}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+\dots+\frac{2}{2^n}-\(\frac{2n-1}{2^{n+1}}\)
= \\
=\frac{1}{2}-\(\frac{2n-1}{2^{n+1}}\)+2\cdot\(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots+\frac{1}{2^n}\)~~~~~*eq_3[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\bf S(soma)~~DO~~ENUNCIADO: \\S = \frac{1}{2}+ \frac{3}{2^2}+ \frac{5}{2^3}+\frac{7}{2^4}\dots +\frac{2n-3}{2^{n-1}}+\frac{2n-1}{2^n} ~~~~~~*eq_1\\\ \\ \\\ \\ \\
MULTIPLICANDO ~~S~~ POR~~\frac{1}{2}: \\S\cdot \frac{1}{2} = \frac{1}{2^2}+ \frac{3}{2^3}+ \frac{5}{2^4}\dots +\frac{2n-3}{2^{n}}+\frac{2n-1}{2^{n+1}}~~~~~~*eq_2 \\\ \\ \\\ \\ \\ SUBTRAINDO~~AS~~DUAS~~EQUAÇÕES(eq_1-eq_2): \\
S-S\cdot\frac{1}{2} = \frac{1}{2}+\frac{3}{2^2}-\frac{1}{2^2}+\frac{5}{2^3}-\frac{3}{2^3}+\frac{7}{2^4}-\frac{5}{2^4} \dots +\frac{2n-1}{2^n}-\frac{2n-3}{2^n}-\(\frac{2n-1}{2^{n+1}}\) = \\
=\frac{1}{2}+\frac{1}{2^2}\cdot(3-1)+\frac{1}{2^3}\cdot(5-3)+\frac{1}{2^4}\cdot(7-5)+\dots+\frac{1}{2^n}\cdot(2n-1-2n+3)-\(\frac{2n-1}{2^{n+1}}\) = \\
=\frac{1}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+\dots+\frac{2}{2^n}-\(\frac{2n-1}{2^{n+1}}\)
= \\
=\frac{1}{2}-\(\frac{2n-1}{2^{n+1}}\)+2\cdot\(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots+\frac{1}{2^n}\)~~~~~*eq_3[/tex3]

Pela soma de uma PG com q<1([tex3]\frac{1}{2}[/tex3]), temos:
[tex3]\bf \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots+\frac{1}{2^n} =\frac{\frac{1}{2}\cdot (\frac{1}{2}^n-1)}{\frac{1}{2}-1}=1-\frac{1}{2^n} ~~~~~~~~~*eq_4[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\bf \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots+\frac{1}{2^n} =\frac{\frac{1}{2}\cdot (\frac{1}{2}^n-1)}{\frac{1}{2}-1}=1-\frac{1}{2^n} ~~~~~~~~~*eq_4[/tex3]

(pegando termo da eq3 e aplicando eq4): [tex3]\bf 2\cdot\(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots+\frac{1}{2^n}\) = 2\cdot \(1-\frac{1}{2^n}-\frac{1}{2}\)=\bf1-\frac{1}{2^{n-1}}~~~~~~~*eq_5[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\bf 2\cdot\(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots+\frac{1}{2^n}\) = 2\cdot \(1-\frac{1}{2^n}-\frac{1}{2}\)=\bf1-\frac{1}{2^{n-1}}~~~~~~~*eq_5[/tex3]

Substitindo eq5 na eq3:
[tex3]\bf S-S\cdot\frac{1}{2} =\frac{1}{2}-\(\frac{2n-1}{2^{n+1}}\)+1-\frac{1}{2^{n-1}} = \\
S\cdot (1-\frac{1}{2})=\frac{1}{2}-\(\frac{2n-1}{2^{n+1}}\)+1-\frac{1}{2^{n-1}} \rightarrow \\
S = 1+\frac{1-2n}{2^n}+2-\frac{1}{2^{n-2}} = 3+\frac{1-2n}{2^n}-\frac{2^2}{2^{n}}=3+\frac{-3-2n}{2^n}=\bf 3-\frac{3+2n}{2^n}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\bf S-S\cdot\frac{1}{2} =\frac{1}{2}-\(\frac{2n-1}{2^{n+1}}\)+1-\frac{1}{2^{n-1}} = \\
S\cdot (1-\frac{1}{2})=\frac{1}{2}-\(\frac{2n-1}{2^{n+1}}\)+1-\frac{1}{2^{n-1}} \rightarrow \\
S = 1+\frac{1-2n}{2^n}+2-\frac{1}{2^{n-2}} = 3+\frac{1-2n}{2^n}-\frac{2^2}{2^{n}}=3+\frac{-3-2n}{2^n}=\bf 3-\frac{3+2n}{2^n}[/tex3]