∑
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}}}
Soma simples
∑
i
=
1
n
x
i
=
x
1
+
x
2
+
.
.
.
+
x
n
{\displaystyle \sum _{i=1}^{n}x_{i}=x_{1}+x_{2}+...+x_{n}}
Soma de quadrados
∑
i
=
1
n
x
i
2
=
x
1
2
+
x
2
2
+
.
.
.
+
x
n
2
{\displaystyle \sum _{i=1}^{n}x_{i}^{2}=x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}}
Quadrado da soma
(
∑
i
=
1
n
x
i
)
2
=
(
x
1
+
x
2
+
.
.
.
+
x
n
)
2
{\displaystyle (\sum _{i=1}^{n}x_{i})^{2}=(x_{1}+x_{2}+...+x_{n})^{2}}
Soma de produtos
∑
i
=
1
n
x
i
y
i
=
x
1
y
1
+
x
2
y
2
+
.
.
.
+
x
n
y
n
{\displaystyle \sum _{i=1}^{n}x_{i}y_{i}=x_{1}y_{1}+x_{2}y_{2}+...+x_{n}y_{n}}
Produtos das somas
(
∑
i
=
1
n
x
i
)
(
∑
j
=
1
m
y
j
)
=
(
x
1
+
x
2
+
.
.
.
+
x
n
)
(
y
1
+
y
2
+
.
.
.
+
y
n
)
{\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})}
Multiplicação por constante
∑
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.
Passo base: s = t
∑
n
=
s
t
C
⋅
f
(
n
)
=
C
⋅
f
(
n
)
{\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
k
∈
N
,
k
>
s
{\displaystyle k\in N,k>s}
arbitrário:
∑
n
=
s
k
C
⋅
f
(
n
)
=
C
⋅
∑
n
=
s
k
f
(
n
)
{\displaystyle \sum _{n=s}^{k}C\cdot f(n)=C\cdot \sum _{n=s}^{k}f(n)}
(Hipótese de indução)
Para
k
+
1
{\displaystyle k+1}
, assumindo o lado esquerdo da equação, temos:
∑
n
=
s
k
+
1
C
⋅
f
(
n
)
=
C
⋅
f
(
k
+
1
)
+
∑
n
=
s
k
C
⋅
f
(
n
)
{\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:
∑
n
=
s
k
+
1
C
⋅
f
(
n
)
=
C
⋅
f
(
k
+
1
)
+
C
⋅
∑
n
=
s
k
f
(
n
)
{\displaystyle \sum _{n=s}^{k+1}C\cdot f(n)=C\cdot f(k+1)+C\cdot \sum _{n=s}^{k}f(n)}
Expandindo
k
−
s
{\displaystyle k-s}
vezes:
∑
n
=
s
k
+
1
C
⋅
f
(
n
)
=
C
⋅
(
f
(
k
+
1
)
)
+
C
⋅
(
f
(
k
)
+
f
(
k
−
1
)
+
.
.
.
+
f
(
s
+
1
)
+
f
(
s
)
)
{\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
C
{\displaystyle C}
em evidência:
∑
n
=
s
k
+
1
C
⋅
f
(
n
)
=
C
⋅
(
f
(
k
+
1
)
+
f
(
k
)
+
f
(
k
−
1
)
+
.
.
.
+
f
(
s
+
1
)
+
f
(
s
)
)
{\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))}
∑
n
=
s
k
+
1
C
⋅
f
(
n
)
=
C
⋅
∑
n
=
s
k
+
1
f
(
n
)
{\displaystyle \sum _{n=s}^{k+1}C\cdot f(n)=C\cdot \sum _{n=s}^{k+1}f(n)}
Portanto:
∑
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,
∀
s
,
t
∈
N
{\displaystyle \forall s,t\in N}
.
Mudança de índices
∑
n
=
s
t
f
(
n
)
=
∑
n
=
s
+
1
t
+
1
f
(
n
−
1
)
{\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=s+1}^{t+1}f(n-1)}
Passo base: s = t
∑
n
=
s
t
f
(
n
)
=
f
(
n
)
=
∑
n
=
s
+
1
t
+
1
f
(
n
−
1
)
{\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
k
∈
N
,
k
>
s
{\displaystyle k\in N,k>s}
arbitrário:
∑
n
=
s
k
f
(
n
)
=
∑
n
=
s
+
1
k
+
1
f
(
n
−
1
)
{\displaystyle \sum _{n=s}^{k}f(n)=\sum _{n=s+1}^{k+1}f(n-1)}
(Hipótese de indução)
Para
k
+
1
{\displaystyle k+1}
, assumindo o lado esquerdo da equação, temos:
∑
n
=
s
k
+
1
f
(
n
)
=
f
(
k
+
1
)
+
∑
n
=
s
k
f
(
n
)
{\displaystyle \sum _{n=s}^{k+1}f(n)=f(k+1)+\sum _{n=s}^{k}f(n)}
, pela definição de somatório.
Aplicando a HI:
∑
n
=
s
k
+
1
f
(
n
)
=
f
(
k
+
1
)
+
∑
n
=
s
+
1
k
+
1
f
(
n
−
1
)
{\displaystyle \sum _{n=s}^{k+1}f(n)=f(k+1)+\sum _{n=s+1}^{k+1}f(n-1)}
Expandindo
k
−
s
{\displaystyle k-s}
vezes:
∑
n
=
s
k
+
1
f
(
n
)
=
f
(
k
+
1
)
+
f
(
k
+
1
−
1
)
+
f
(
k
−
1
)
+
.
.
.
+
f
(
s
−
1
)
+
f
(
s
+
1
−
1
)
{\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)}
∑
n
=
s
k
+
1
f
(
n
)
=
f
(
k
+
1
)
+
f
(
k
)
+
f
(
k
−
1
)
+
.
.
.
+
f
(
s
−
1
)
+
f
(
s
)
{\displaystyle \sum _{n=s}^{k+1}f(n)=f(k+1)+f(k)+f(k-1)+...+f(s-1)+f(s)}
∑
n
=
s
k
+
1
f
(
n
)
=
∑
n
=
s
+
1
k
+
2
f
(
n
−
1
)
{\displaystyle \sum _{n=s}^{k+1}f(n)=\sum _{n=s+1}^{k+2}f(n-1)}
, uma vez que existem
k
+
2
{\displaystyle k+2}
termos.
Portanto:
∑
n
=
s
t
f
(
n
)
=
∑
n
=
s
+
1
t
+
1
f
(
n
−
1
)
∀
s
,
t
∈
N
{\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=s+1}^{t+1}f(n-1)\forall s,t\in N}
.