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== 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=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>
<math> \sum^n_{i = m} f(i) = \sum^{n+p}_{i = m+p} f(i-p) </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_{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=0}^n i = \sum_{i=1}^n i = \frac{n(n+1)}{2} </math>
<math> \sum\limits_{k=0}^{n-1}{2^k} = 2^n-1 </math>
<math> \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 </math>
<math> \sum_{i=0}^n i^3 = \left(\sum_{i=0}^n i\right)^2 </math>
<math> \sum_{i=m}^{n-1} a^i = \frac{a^m-a^n}{1-a} (m < n) </math>
<math> \sum_{i=0}^{n-1} a^i = \frac{1-a^n}{1-a} </math>
Somatório e Produtório (view source)
Revision as of 12:28, 25 November 2015
, 12:28, 25 November 2015Created 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=..."
== 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=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>
<math> \sum^n_{i = m} f(i) = \sum^{n+p}_{i = m+p} f(i-p) </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_{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=0}^n i = \sum_{i=1}^n i = \frac{n(n+1)}{2} </math>
<math> \sum\limits_{k=0}^{n-1}{2^k} = 2^n-1 </math>
<math> \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 </math>
<math> \sum_{i=0}^n i^3 = \left(\sum_{i=0}^n i\right)^2 </math>
<math> \sum_{i=m}^{n-1} a^i = \frac{a^m-a^n}{1-a} (m < n) </math>
<math> \sum_{i=0}^{n-1} a^i = \frac{1-a^n}{1-a} </math>
