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(Created page with " == Propriedades de Somatório == <math> \sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n)\text{, onde C é uma constante. </math> <math> \sum_{n=s}^t f(n) + \sum_{n=...")
 
 
(46 intermediate revisions by 3 users not shown)
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== Propriedades de Somatório ==
 
== Propriedades de Somatório ==
  
  
−
 
+
<math> \sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n) </math>, onde C é uma constante.  
−
 
 
−
<math> \sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n)\text{, onde C é uma constante. </math>
 
  
 
<math> \sum_{n=s}^t f(n) + \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) + g(n)\right] </math>
 
<math> \sum_{n=s}^t f(n) + \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) + g(n)\right] </math>
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<math> \sum\limits_{n=s}^{t} j  =  \sum\limits_{n=1}^{t} j - \sum\limits_{n=1}^{s-1} j  </math>
 
<math> \sum\limits_{n=s}^{t} j  =  \sum\limits_{n=1}^{t} j - \sum\limits_{n=1}^{s-1} j  </math>
  
−
<math> \sum_{n=s}^j f(n) + \sum_{n=j+1}^t f(n) = \sum_{n=s}^t f(n) \text{, note que } s \leq  j \leq t </math>
+
<math> \sum_{n=s}^j f(n) + \sum_{n=j+1}^t f(n) = \sum_{n=s}^t f(n)</math>, note que <math>  s \leq  j \leq t </math>
  
−
<math> \sum_{i=m}^n i = \frac{n(n+1)}{2} - \frac{m(m-1)}{2} = \frac{(n+1-m)(n+m)}{2}, \text{ progressão aritmética.} </math>
+
<math> \sum_{i=m}^n i = \frac{n(n+1)}{2} - \frac{m(m-1)}{2} = \frac{(n+1-m)(n+m)}{2},</math> progressão aritmética.
  
 
<math> \sum_{i=0}^n i = \sum_{i=1}^n i = \frac{n(n+1)}{2} </math>
 
<math> \sum_{i=0}^n i = \sum_{i=1}^n i = \frac{n(n+1)}{2} </math>
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<math> \sum_{i=0}^{n-1} a^i = \frac{1-a^n}{1-a} </math>
 
<math> \sum_{i=0}^{n-1} a^i = \frac{1-a^n}{1-a} </math>
 +
 +
 +
----
 +
 +
== Principais representações ==
 +
====Soma simples====
 +
<math>\sum_{i=1}^{n} x_i = x_1+x_2+...+x_n</math>
 +
 +
====Soma de quadrados====
 +
<math>\sum_{i=1}^{n} x_i^2 = x_1^2+x_2^2+...+x_n^2</math>
 +
 +
====Quadrado da soma====
 +
<math>(\sum_{i=1}^{n} x_i)^2 = (x_1+x_2+...+x_n)^2</math>
 +
 +
====Soma de produtos====
 +
<math>\sum_{i=1}^{n} x_iy_i = x_1y_1+x_2y_2+...+x_ny_n</math>
 +
 +
====Produtos das somas====
 +
<math>(\sum_{i=1}^{n} x_i)(\sum_{j=1}^{m} y_j) = (x_1+x_2+...+x_n)(y_1+y_2+...+y_n)</math>
 +
 +
----
 +
 +
== Aplicação das Propriedades ==
 +
Alguns exemplos de aplicações das propriedades do somatório:
 +
 +
=== Exemplo 1 ===
 +
 +
Utilize  as propriedades de notação  de  somatório e,
 +
possivelmente, mudança  de índice  para deduzir que
 +
<math>\sum_{j=1}^n (a_j - a_{j-1})</math> é igual a <math>a_n - a_0</math>,
 +
onde <math>(a_i )_{i=0}^{\infty}</math> é  uma sequência  de números  reais.
 +
Este  tipo de  soma  é bastante conhecida em Matemática como ''soma telescópica''.
 +
 +
==== Resolução ====
 +
 +
<math> \sum_{j=1}^n (a_j - a_{j-1}) = (a_n - a_{n-1}) + \sum_{j=1}^{n-1}</math>
 +
 +
 +
Expandindo <math>n</math> vezes:
 +
 +
<math> \sum_{j=1}^n (a_j - a_{j-1}) = (a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + ... + (a_2 - a_1) + (a_1 - a_0)</math>
 +
 +
<math> \sum_{j=1}^n (a_j - a_{j-1}) = a_n - \cancel{a_{n-1}} + \cancel{a_{n-1}} - \cancel{a_{n-2}} + ... + \cancel{a_2} - \cancel{a_1} + \cancel{a_1} - a_0</math>
 +
 +
<math> \sum_{j=1}^n (a_j - a_{j-1}) = a_n - a_0 </math>
 +
 +
=== Exemplo 2 ===
 +
 +
O objetivo deste problema é encontrar uma fórmula fechada para
 +
 +
<math>\sum_{k=1}^n (2 \cdot k - 1 )</math>
 +
 +
 +
Para tal, note que
 +
 +
<math>k^2 - ( k - 1)^2 = 2k - 1</math>
 +
 +
 +
Logo,
 +
 +
<math>\sum_{k=1}^n \left ( k^2 - ( k - 1)^2 \right ) = \sum_{k=1}^n ( 2k - 1) </math>
 +
 +
 +
Então, utilize o resultado do problema conhecido como  "soma telescópia" do exemplo 1 para encontrar  a fórmula
 +
desejada.
 +
 +
==== Resolução ====
 +
 +
<math> \sum_{k=1}^n 2 \cdot k - 1 = \sum_{k=1}^n k^2 - (k-1)^2</math>
 +
 +
 +
Pela fórmula da soma telescópica
 +
 +
<math> \sum_{k=1}^n 2 \cdot k - 1 = n^2 - 0^2 = n^2</math>
 +
 +
=== Exemplo 3 ===
 +
 +
Utilize as  propriedades de notação de somatório e os seus  conhecimentos de soma  de termos de uma PA para
 +
calcular
 +
 +
<math>\sum_{k=1}^n (2 \cdot k - 1 )</math>
 +
 +
de forma  distinta daquela usada no problema anterior. Qual  das duas
 +
soluções lhe parece mais fácil?
 +
 +
==== Resolução ====
 +
 +
<math>\sum_{k=1}^n (2 \cdot k - 1 ) = \sum_{k=1}^n 2 \cdot k - \sum_{k=1}^n 1</math>
 +
 +
<math>\sum_{k=1}^n (2 \cdot k - 1 ) = 2 \cdot \sum_{k=1}^n k - n</math>
 +
 +
<math>\sum_{k=1}^n (2 \cdot k - 1 ) = 2 \cdot \frac{n \cdot (n+1)}{2} - n</math>
 +
 +
<math>\sum_{k=1}^n (2 \cdot k - 1 ) = n^2 + n - n</math>
 +
 +
<math>\sum_{k=1}^n (2 \cdot k - 1 ) = n^2</math>
 +
 +
===Exemplo 4===
 +
Suprimindo um dos elementos do conjunto {<math>1, 2, . . . , n</math>}, a média aritmética dos elementos
 +
 +
16,1. Determine o valor de n e qual foi o elemento suprimido do conjunto para o cálculo da média.
 +
 +
==== Resolução ====
 +
<math>\sum_{k=1}^{n} k = 1+2+...+n= \frac{n \cdot (n+1)}{2}</math>
 +
 +
média aritmética é dada por :
 +
 +
<math>\frac {\frac{n \cdot (n+1)}{2}}{n} = \frac{n \cdot (n+1)}{2n} = \frac{1}{n}\sum_{k=1}^{n} k</math>
 +
 +
Pela propriedade da progressão aritmética
 +
 +
 +
<math>\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}</math>
 +
 +
 +
usando a função de calculo da média:
 +
 +
<math>\frac{(n-1)n}{2(n-1)} = \frac{n}{2} = 16,1</math>
 +
 +
<math>n = 32,2</math>
 +
 +
Substituindo <math>n</math> na equação:
 +
 +
<math>n-1 = 31,2</math>
 +
 +
<math>\sum_{k=1}^{n-1} k = 517,92</math>
 +
 +
<math>\sum_{k=1}^{n} k = 534,52</math>
 +
 +
Portanto o termo omitido foi:
 +
 +
<math>534,52 - 517,92 = 16,6</math>
 +
 +
===Exemplo 5===
 +
Encontre uma fórmula fechada
 +
 +
<math>\sum_{k=1}^{n} k^3</math>
 +
 +
onde <math>n \in N \text{, com } n \geq 1</math> .
 +
==== Resolução ====
 +
<math>\sum_{k=1}^{n} k^3 = \sum_{k=1}^{n} kk^2 = \sum_{k=1}^{n} k \sum_{k=1}^{n} k^2</math>
 +
 +
Temos:
 +
 +
 +
<pre>Incompleto
 +
</pre>
 +
 +
===Exemplo 6===
 +
Calcule a soma
 +
 +
<math>\sum_{k=1}^{n} k\cdot k!</math>
 +
 +
onde <math> n \in N \text{,com } n \geq 1</math>
 +
 +
==== Resolução ====
 +
Separando o somatório:
 +
 +
<math>\sum_{k=1}^{n} k\cdot k! =\sum_{k=1}^{n} k\cdot \sum_{k=1}^{n} k!  </math>/
 +
 +
teremos que descobrir o
 +
 +
<math>\sum_{k=1}^{n} k!</math>
 +
 +
então
 +
 +
<math>\sum_{k=1}^{n} k!+(n+1)! = \sum_{k=0}^{n} (k!+1)! </math>
 +
 +
<math>1+\sum_{k=1}^{n} (k+1)! = 1+\sum_{k=1}^{n} (k+1)k!</math>
 +
 +
<pre>Incompleto
 +
</pre>
 +
 +
===Exemplo 7===
 +
Os números <math>\sqrt{2}, \quad \sqrt{3} \quad \text{e} \quad \sqrt{5}</math>
 +
 +
podem pertencer a uma mesma progressão aritmética?
 +
 +
==== Resolução ====
 +
Assumindo uma PA  <math>(\sqrt{1},\sqrt{2}, \sqrt{3}, \sqrt{4},\sqrt{5})</math>
 +
 +
os termos <math>\sqrt{2}, \quad \sqrt{3} \quad \text{e} \quad \sqrt{5}</math>
 +
pertencem a essa progressão se pela propriedade da progressão aritmética a média aritmética dos termos da ponta de uma sequencia (a, b e c)
 +
for igual a o termo do meio:
 +
 +
<math>\frac {a+c}{2}= b </math>
 +
 +
<math>\sqrt{3}\simeq1,7</math>
 +
 +
inserindo os valores na equação:
 +
<math>\frac {\sqrt{1}+\sqrt{5}}{2}\simeq1,6 </math>
 +
 +
<math>\frac {\sqrt{2}+\sqrt{4}}{2}\simeq 1,7 </math>
 +
 +
<math>\frac {\sqrt{2}+\sqrt{5}}{2} \simeq 1,8 </math>
 +
 +
 +
Portanto <math>\sqrt{2}, \quad \sqrt{3} \quad \text{e} \quad \sqrt{5}</math> não pertencem a mesma progressão aritmética.
 +
 +
 +
----
 +
 +
== Provas de algumas propriedades ==
 +
===Multiplicação por constante===
 +
<math> \sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n) </math>, onde C é uma constante.
 +
 +
===== Passo base: s = t =====
 +
<math> \sum_{n=s}^t C\cdot f(n) = C\cdot f(n) </math>, pela definição de somatório.
 +
 +
===== Passo indutivo: s < t =====
 +
 +
Suponha que para um <math>k \in N, k > s</math> arbitrário:
 +
 +
<math> \sum_{n=s}^k C\cdot f(n) = C\cdot \sum_{n=s}^k f(n) </math> (Hipótese de indução)
 +
 +
 +
Para <math>k+1</math>, assumindo o lado esquerdo da equação, temos:
 +
 +
<math> \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot f(k+1) + \sum_{n=s}^k C\cdot f(n)</math>, pela definição de somatório.
 +
 +
 +
Aplicando a HI:
 +
 +
<math> \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot f(k+1) + C\cdot \sum_{n=s}^k f(n)</math>
 +
 +
 +
Expandindo <math>k-s</math> vezes:
 +
 +
<math> \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot (f(k+1)) + C\cdot (f(k) + f(k-1) + ... + f(s+1) + f(s))</math>
 +
 +
 +
Colocando <math>C</math> em evidência:
 +
 +
<math> \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot (f(k+1) + f(k) + f(k-1) + ... + f(s+1) + f(s))</math>
 +
 +
<math> \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot \sum_{n=s}^{k+1} f(n) </math>
 +
 +
 +
Portanto:
 +
 +
<math> \sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n) </math>, onde C é uma constante, <math>\forall s, t \in N</math>.
 +
 +
 +
 +
=== Mudança de índices ===
 +
<math> \sum_{n=s}^t f(n) = \sum_{n=s+1}^{t+1} f(n-1) </math>
 +
 +
===== Passo base: s = t =====
 +
<math> \sum_{n=s}^t f(n) = f(n) = \sum_{n=s+1}^{t+1} f(n-1) </math>, pela definição de somatório.
 +
 +
===== Passo indutivo: s < t =====
 +
 +
Suponha que para um <math>k \in N, k > s</math> arbitrário:
 +
 +
<math> \sum_{n=s}^k f(n) = \sum_{n=s+1}^{k+1} f(n-1) </math> (Hipótese de indução)
 +
 +
 +
Para <math>k+1</math>, assumindo o lado esquerdo da equação, temos:
 +
 +
<math> \sum_{n=s}^{k+1} f(n) = f(k+1) + \sum_{n=s}^k f(n)</math>, pela definição de somatório.
 +
 +
 +
Aplicando a HI:
 +
 +
<math> \sum_{n=s}^{k+1} f(n) = f(k+1) + \sum_{n=s+1}^{k+1} f(n-1)</math>
 +
 +
 +
Expandindo <math>k-s</math> vezes:
 +
 +
<math> \sum_{n=s}^{k+1} f(n) = f(k+1) + f(k+1-1) + f(k-1) + ... + f(s-1) + f(s+1-1)</math>
 +
 +
<math> \sum_{n=s}^{k+1} f(n) = f(k+1) + f(k) + f(k-1) + ... + f(s-1) + f(s)</math>
 +
 +
<math> \sum_{n=s}^{k+1} f(n) = \sum_{n=s+1}^{k+2} f(n-1)</math>, uma vez que existem <math>k+2</math> termos.
 +
 +
 +
Portanto:
 +
 +
<math> \sum_{n=s}^t f(n) = \sum_{n=s+1}^{t+1} f(n-1) \forall s, t \in N</math>.
 +
----
 +
 +
== Somatório em Linguagem Funcional ==
 +
 +
====Elixir<ref>https://github.com/jaimerson/fmc-elixir-somatorio</ref>====
 +
<pre>
 +
defmodule FMC do
 +
  def somatorio(start \\0, finish, callback)
 +
 +
  def somatorio(start, finish, callback) when start == finish do
 +
    callback.(start)
 +
  end
 +
 +
  def somatorio(start, finish, callback) do
 +
    _somatorio(Enum.to_list(start..finish), callback)
 +
  end
 +
 +
  defp _somatorio([], _), do: 0
 +
  defp _somatorio([head | tail], callback) do
 +
    callback.(head) + _somatorio(tail, callback)
 +
  end
 +
end
 +
</pre>
 +
 +
----
 +
 +
==Referências==
 +
<references />
 +
----
 +
==Autores==
 +
<pre>Jaimerson Araújo
 +
 +
Francleide Simão
 +
</pre>

Latest revision as of 09:55, 10 December 2015

Propriedades de Somatório

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n) } , onde C é uma constante.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^t f(n) + \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) + g(n)\right] }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^t f(n) - \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) - g(n)\right] }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum^n_{i = m} f(i) = \sum^{n+p}_{i = m+p} f(i-p) }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum\limits_{n=s}^{t} j = \sum\limits_{n=1}^{t} j - \sum\limits_{n=1}^{s-1} j }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^j f(n) + \sum_{n=j+1}^t f(n) = \sum_{n=s}^t f(n)} , note que Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s \leq j \leq t }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=m}^n i = \frac{n(n+1)}{2} - \frac{m(m-1)}{2} = \frac{(n+1-m)(n+m)}{2},} progressão aritmética.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=0}^n i = \sum_{i=1}^n i = \frac{n(n+1)}{2} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum\limits_{k=0}^{n-1}{2^k} = 2^n-1 }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=s}^m\sum_{j=t}^n {a_i}{c_j} = \sum_{i=s}^m a_i \cdot \sum_{j=t}^n c_j }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=0}^n i^3 = \left(\sum_{i=0}^n i\right)^2 }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=m}^{n-1} a^i = \frac{a^m-a^n}{1-a} (m < n) }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=0}^{n-1} a^i = \frac{1-a^n}{1-a} }


Principais representações

Soma simples

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^{n} x_i = x_1+x_2+...+x_n}

Soma de quadrados

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^{n} x_i^2 = x_1^2+x_2^2+...+x_n^2}

Quadrado da soma

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\sum_{i=1}^{n} x_i)^2 = (x_1+x_2+...+x_n)^2}

Soma de produtos

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^{n} x_iy_i = x_1y_1+x_2y_2+...+x_ny_n}

Produtos das somas

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\sum_{i=1}^{n} x_i)(\sum_{j=1}^{m} y_j) = (x_1+x_2+...+x_n)(y_1+y_2+...+y_n)}

Aplicação das Propriedades

Alguns exemplos de aplicações das propriedades do somatório:

Exemplo 1

Utilize as propriedades de notação de somatório e, possivelmente, mudança de índice para deduzir que Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{j=1}^n (a_j - a_{j-1})} é igual a Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n - a_0} , onde Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_i )_{i=0}^{\infty}} é uma sequência de números reais. Este tipo de soma é bastante conhecida em Matemática como soma telescópica.

Resolução

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{j=1}^n (a_j - a_{j-1}) = (a_n - a_{n-1}) + \sum_{j=1}^{n-1}}


Expandindo Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} vezes:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{j=1}^n (a_j - a_{j-1}) = (a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + ... + (a_2 - a_1) + (a_1 - a_0)}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{j=1}^n (a_j - a_{j-1}) = a_n - \cancel{a_{n-1}} + \cancel{a_{n-1}} - \cancel{a_{n-2}} + ... + \cancel{a_2} - \cancel{a_1} + \cancel{a_1} - a_0}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{j=1}^n (a_j - a_{j-1}) = a_n - a_0 }

Exemplo 2

O objetivo deste problema é encontrar uma fórmula fechada para

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n (2 \cdot k - 1 )}


Para tal, note que

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k^2 - ( k - 1)^2 = 2k - 1}


Logo,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n \left ( k^2 - ( k - 1)^2 \right ) = \sum_{k=1}^n ( 2k - 1) }


Então, utilize o resultado do problema conhecido como "soma telescópia" do exemplo 1 para encontrar a fórmula desejada.

Resolução

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n 2 \cdot k - 1 = \sum_{k=1}^n k^2 - (k-1)^2}


Pela fórmula da soma telescópica

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n 2 \cdot k - 1 = n^2 - 0^2 = n^2}

Exemplo 3

Utilize as propriedades de notação de somatório e os seus conhecimentos de soma de termos de uma PA para calcular

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n (2 \cdot k - 1 )}

de forma distinta daquela usada no problema anterior. Qual das duas soluções lhe parece mais fácil?

Resolução

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n (2 \cdot k - 1 ) = \sum_{k=1}^n 2 \cdot k - \sum_{k=1}^n 1}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n (2 \cdot k - 1 ) = 2 \cdot \sum_{k=1}^n k - n}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n (2 \cdot k - 1 ) = 2 \cdot \frac{n \cdot (n+1)}{2} - n}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n (2 \cdot k - 1 ) = n^2 + n - n}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^n (2 \cdot k - 1 ) = n^2}

Exemplo 4

Suprimindo um dos elementos do conjunto {Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1, 2, . . . , n} }, a média aritmética dos elementos

16,1. Determine o valor de n e qual foi o elemento suprimido do conjunto para o cálculo da média.

Resolução

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n} k = 1+2+...+n= \frac{n \cdot (n+1)}{2}}

média aritmética é dada por :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {\frac{n \cdot (n+1)}{2}}{n} = \frac{n \cdot (n+1)}{2n} = \frac{1}{n}\sum_{k=1}^{n} k}

Pela propriedade da progressão aritmética


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}}


usando a função de calculo da média:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{(n-1)n}{2(n-1)} = \frac{n}{2} = 16,1}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n = 32,2}

Substituindo Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} na equação:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n-1 = 31,2}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n-1} k = 517,92}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n} k = 534,52}

Portanto o termo omitido foi:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 534,52 - 517,92 = 16,6}

Exemplo 5

Encontre uma fórmula fechada

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n} k^3}

onde Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in N \text{, com } n \geq 1} .

Resolução

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n} k^3 = \sum_{k=1}^{n} kk^2 = \sum_{k=1}^{n} k \sum_{k=1}^{n} k^2}

Temos: 


Incompleto

Exemplo 6

Calcule a soma

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n} k\cdot k!}

onde Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in N \text{,com } n \geq 1}

Resolução

Separando o somatório:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n} k\cdot k! =\sum_{k=1}^{n} k\cdot \sum_{k=1}^{n} k! } /

teremos que descobrir o

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n} k!}

então

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{n} k!+(n+1)! = \sum_{k=0}^{n} (k!+1)! }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1+\sum_{k=1}^{n} (k+1)! = 1+\sum_{k=1}^{n} (k+1)k!} 

Incompleto

Exemplo 7

Os números Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{2}, \quad \sqrt{3} \quad \text{e} \quad \sqrt{5}}

podem pertencer a uma mesma progressão aritmética?

Resolução

Assumindo uma PA Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\sqrt{1},\sqrt{2}, \sqrt{3}, \sqrt{4},\sqrt{5})}

os termos Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{2}, \quad \sqrt{3} \quad \text{e} \quad \sqrt{5}} pertencem a essa progressão se pela propriedade da progressão aritmética a média aritmética dos termos da ponta de uma sequencia (a, b e c) for igual a o termo do meio:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {a+c}{2}= b }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{3}\simeq1,7}

inserindo os valores na equação: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {\sqrt{1}+\sqrt{5}}{2}\simeq1,6 }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {\sqrt{2}+\sqrt{4}}{2}\simeq 1,7 }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {\sqrt{2}+\sqrt{5}}{2} \simeq 1,8 }


Portanto Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{2}, \quad \sqrt{3} \quad \text{e} \quad \sqrt{5}} não pertencem a mesma progressão aritmética.


Provas de algumas propriedades

Multiplicação por constante

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n) } , onde C é uma constante.

Passo base: s = t

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^t C\cdot f(n) = C\cdot f(n) } , pela definição de somatório.

Passo indutivo: s < t

Suponha que para um Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in N, k > s} arbitrário:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^k C\cdot f(n) = C\cdot \sum_{n=s}^k f(n) } (Hipótese de indução)


Para Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k+1} , assumindo o lado esquerdo da equação, temos:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot f(k+1) + \sum_{n=s}^k C\cdot f(n)} , pela definição de somatório.


Aplicando a HI:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot f(k+1) + C\cdot \sum_{n=s}^k f(n)}


Expandindo Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k-s} vezes:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot (f(k+1)) + C\cdot (f(k) + f(k-1) + ... + f(s+1) + f(s))}


Colocando Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} em evidência:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot (f(k+1) + f(k) + f(k-1) + ... + f(s+1) + f(s))}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} C\cdot f(n) = C\cdot \sum_{n=s}^{k+1} f(n) }


Portanto:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n) } , onde C é uma constante, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \forall s, t \in N} .


Mudança de índices

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^t f(n) = \sum_{n=s+1}^{t+1} f(n-1) }

Passo base: s = t

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^t f(n) = f(n) = \sum_{n=s+1}^{t+1} f(n-1) } , pela definição de somatório.

Passo indutivo: s < t

Suponha que para um Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in N, k > s} arbitrário:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^k f(n) = \sum_{n=s+1}^{k+1} f(n-1) } (Hipótese de indução)


Para Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k+1} , assumindo o lado esquerdo da equação, temos:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} f(n) = f(k+1) + \sum_{n=s}^k f(n)} , pela definição de somatório.


Aplicando a HI:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} f(n) = f(k+1) + \sum_{n=s+1}^{k+1} f(n-1)}


Expandindo Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k-s} vezes:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} f(n) = f(k+1) + f(k+1-1) + f(k-1) + ... + f(s-1) + f(s+1-1)}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} f(n) = f(k+1) + f(k) + f(k-1) + ... + f(s-1) + f(s)}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^{k+1} f(n) = \sum_{n=s+1}^{k+2} f(n-1)} , uma vez que existem Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k+2} termos.


Portanto:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=s}^t f(n) = \sum_{n=s+1}^{t+1} f(n-1) \forall s, t \in N} .

Somatório em Linguagem Funcional

Elixir[1]

defmodule FMC do
  def somatorio(start \\0, finish, callback)

  def somatorio(start, finish, callback) when start == finish do
    callback.(start)
  end

  def somatorio(start, finish, callback) do
    _somatorio(Enum.to_list(start..finish), callback)
  end

  defp _somatorio([], _), do: 0
  defp _somatorio([head | tail], callback) do
    callback.(head) + _somatorio(tail, callback)
  end
end

Referências

  1. Jump up ↑ https://github.com/jaimerson/fmc-elixir-somatorio

Autores

Jaimerson Araújo

Francleide Simão
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