Question

Resolva o Somatório:

$\large \sf \: \displaystyle \large \sf \:\sum^{4}_{n = 0} \dfrac{ {( - 1)}^{n} }{1 + n}$

Letra A) 5/7


Letra B) 12/60


Letra C) 10


Letra D) 47/60 ​

Answer 1

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$\Huge\green{\boxed{\rm~~~\red{D)}~\blue{ \dfrac{47}{60} }~~~}}$

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$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

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Grande, Murillão, como estás⁉ Vamos analisar esse exercício❗ Acompanhe a resolução abaixo. ✌

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$\LARGE\gray{\boxed{\sf\blue{~~\displaystyle\sum_{n=0}^{4} \dfrac{(-1)^n}{1 + n}~~}}}$

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☔ Como temos somente 5 termos nesta soma, vamos abri-la e encontrar a solução manualmente (caso fosse uma somatória maior ou até mesmo com infinitos termos tomaríamos uma abordagem diferente):

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$\large\blue{\sf \displaystyle\sum_{n=0}^{4} \dfrac{(-1)^n}{1 + n} =   \dfrac{1}{1} + \dfrac{(-1)}{2} + \dfrac{1}{3} + \dfrac{(-1)}{4} + \dfrac{1}{5}}$

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☔ Sabemos que {1, 2, 3, 4, 5} são primos entre si (com a exceção de {2, 4}), ou seja, o M.M.C. {1, 2, 3, 4, 5} será igual a:

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$\LARGE\blue{\sf \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}{2} = \dfrac{120}{2} = 60 }$

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☔ Retornando para nossa soma teremos:

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$\LARGE\blue{\sf = \dfrac{60}{60} + \dfrac{(-30)}{60} + \dfrac{20}{60} + \dfrac{(-15)}{60} + \dfrac{12}{60}}$

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$\LARGE\blue{\sf = \dfrac{60 - 30 + 20 - 15 + 12}{60} }$

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$\LARGE\blue{\sf = \dfrac{47}{60} }$

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$\Huge\green{\boxed{\rm~~~\red{D)}~\blue{ \dfrac{47}{60} }~~~}}$ ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\Large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$

Answer 2

Resposta:

OLÁ

VAMOS A SUA PERGUNTA:⇒⇒

$\sf \displaystyle\sum _{n=0}^4\frac{\left(-1\right)^n}{1+n}$

$\sf \displaystyle a_n=\frac{\left(-1\right)^n}{1+n}\\\\\\a_0=\dfrac{\left(-1\right)^0}{1+0}\\\\\\\dfrac{\left(-1\right)^0}{1+0}\\\\\\=1$

$\sf \displaystyle a_n=\frac{\left(-1\right)^n}{1+n}\\\\\\a_1=\frac{\left(-1\right)^1}{1+1}\\\\\\\frac{\left(-1\right)^1}{1+1}=-\frac{1}{2} \\\\\\=-\frac{1}{2}$

$\sf \displaystyle a_n=\frac{\left(-1\right)^n}{1+n}\\\\\\a_2=\frac{\left(-1\right)^2}{1+2}\\\\\\\frac{\left(-1\right)^2}{1+2}=\frac{1}{3} \\\\\\=\frac{1}{3}$

$\sf \displaystyle a_n=\frac{\left(-1\right)^n}{1+n}\\\\\\a_3=\frac{\left(-1\right)^3}{1+3}\\\\\\\frac{\left(-1\right)^3}{1+3}=-\frac{1}{4} \\\\\\=-\frac{1}{4}$

$\sf \displaystyle a_n=\frac{\left(-1\right)^n}{1+n}\\\\\\a_4=\frac{\left(-1\right)^4}{1+4}\\\\\\\frac{\left(-1\right)^4}{1+4}=\frac{1}{5} \\\\\\=\frac{1}{5}$

$\sf {Agora \ vc \ calcula \ a \ soma \ de \ \:}\displaystyle a_n=\frac{\left(-1\right)^n}{1+n}{\:para\:}n=0\dots 4$

$\sf \displaystyle =1+\left(-\frac{1}{2}\right)+\frac{1}{3}+\left(-\frac{1}{4}\right)+\frac{1}{5}$

$\sf \sf \displaystyle =1+\left(-\frac{1}{2}\right)+\frac{1}{3}+\left(-\frac{1}{4}\right)+\frac{1}{5}=\frac{47}{60}$

$\sf \boxed{\sf \red{=\dfrac{47}{60}}}$ ← RESPOSTA.

• ≡ LETRA ''D''

Explicação passo-a-passo:

ESPERO TER AJUDADO