∑ n = s t C ⋅ f ( n ) = C ⋅ ∑ n = s t f ( n ) {\displaystyle \sum _{n=s}^{t}C\cdot f(n)=C\cdot \sum _{n=s}^{t}f(n)} onde C é uma constante.
∑ n = s t f ( n ) + ∑ n = s t g ( n ) = ∑ n = s t [ f ( n ) + g ( n ) ] {\displaystyle \sum _{n=s}^{t}f(n)+\sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left[f(n)+g(n)\right]}
∑ n = s t f ( n ) − ∑ n = s t g ( n ) = ∑ n = s t [ f ( n ) − g ( n ) ] {\displaystyle \sum _{n=s}^{t}f(n)-\sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left[f(n)-g(n)\right]}
∑ i = m n f ( i ) = ∑ i = m + p n + p f ( i − p ) {\displaystyle \sum _{i=m}^{n}f(i)=\sum _{i=m+p}^{n+p}f(i-p)}
∑ n = s t j = ∑ n = 1 t j − ∑ n = 1 s − 1 j {\displaystyle \sum \limits _{n=s}^{t}j=\sum \limits _{n=1}^{t}j-\sum \limits _{n=1}^{s-1}j}
∑ n = s j f ( n ) + ∑ n = j + 1 t f ( n ) = ∑ n = s t f ( n ) {\displaystyle \sum _{n=s}^{j}f(n)+\sum _{n=j+1}^{t}f(n)=\sum _{n=s}^{t}f(n)} , note que s ≤ j ≤ t {\displaystyle s\leq j\leq t}
∑ i = m n i = n ( n + 1 ) 2 − m ( m − 1 ) 2 = ( n + 1 − m ) ( n + m ) 2 , {\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.
∑ i = 0 n i = ∑ i = 1 n i = n ( n + 1 ) 2 {\displaystyle \sum _{i=0}^{n}i=\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}}
∑ k = 0 n − 1 2 k = 2 n − 1 {\displaystyle \sum \limits _{k=0}^{n-1}{2^{k}}=2^{n}-1}
∑ i = s m ∑ j = t n a i c j = ∑ i = s m a i ⋅ ∑ j = t n c j {\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}}
∑ i = 0 n i 3 = ( ∑ i = 0 n i ) 2 {\displaystyle \sum _{i=0}^{n}i^{3}=\left(\sum _{i=0}^{n}i\right)^{2}}
∑ i = m n − 1 a i = a m − a n 1 − a ( m < n ) {\displaystyle \sum _{i=m}^{n-1}a^{i}={\frac {a^{m}-a^{n}}{1-a}}(m<n)}
∑ i = 0 n − 1 a i = 1 − a n 1 − a {\displaystyle \sum _{i=0}^{n-1}a^{i}={\frac {1-a^{n}}{1-a}}}
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onde C é uma constante.
![{\displaystyle \sum _{n=s}^{t}f(n)+\sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left[f(n)+g(n)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a762d789e2af544d36eeeefc783b61559d64dc84)
![{\displaystyle \sum _{n=s}^{t}f(n)-\sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left[f(n)-g(n)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3b63ace30b999de1bbdcf04f85ed8c7ce0b7a9a)
, note que 
progressão aritmética.
