∑
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)}
∑
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)}
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}},}
∑
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}}}
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
∑
j
=
1
n
(
a
j
−
a
j
−
1
)
{\displaystyle \sum _{j=1}^{n}(a_{j}-a_{j-1})}
a
n
−
a
0
{\displaystyle a_{n}-a_{0}}
(
a
i
)
i
=
0
∞
{\displaystyle (a_{i})_{i=0}^{\infty }}
soma telescópica .
Resolução
∑
j
=
1
n
(
a
j
−
a
j
−
1
)
=
(
a
n
−
a
n
−
1
)
+
∑
j
=
1
n
−
1
{\displaystyle \sum _{j=1}^{n}(a_{j}-a_{j-1})=(a_{n}-a_{n-1})+\sum _{j=1}^{n-1}}
n
{\displaystyle n}
∑
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
)
{\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})}
∑
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
{\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}}
∑
j
=
1
n
(
a
j
−
a
j
−
1
)
=
a
n
−
a
0
{\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
∑
k
=
1
n
(
2
⋅
k
−
1
)
{\displaystyle \sum _{k=1}^{n}(2\cdot k-1)}
k
2
−
(
k
−
1
)
2
=
2
k
−
1
{\displaystyle k^{2}-(k-1)^{2}=2k-1}
∑
k
=
1
n
(
k
2
−
(
k
−
1
)
2
)
=
∑
k
=
1
n
(
2
k
−
1
)
{\displaystyle \sum _{k=1}^{n}\left(k^{2}-(k-1)^{2}\right)=\sum _{k=1}^{n}(2k-1)}
Resolução
∑
k
=
1
n
2
⋅
k
−
1
=
∑
k
=
1
n
k
2
−
(
k
−
1
)
2
{\displaystyle \sum _{k=1}^{n}2\cdot k-1=\sum _{k=1}^{n}k^{2}-(k-1)^{2}}
∑
k
=
1
n
2
⋅
k
−
1
=
n
2
−
0
2
=
n
2
{\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
∑
k
=
1
n
(
2
⋅
k
−
1
)
{\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
∑
k
=
1
n
(
2
⋅
k
−
1
)
=
∑
k
=
1
n
2
⋅
k
−
∑
k
=
1
n
1
{\displaystyle \sum _{k=1}^{n}(2\cdot k-1)=\sum _{k=1}^{n}2\cdot k-\sum _{k=1}^{n}1}
∑
k
=
1
n
(
2
⋅
k
−
1
)
=
2
⋅
∑
k
=
1
n
k
−
n
{\displaystyle \sum _{k=1}^{n}(2\cdot k-1)=2\cdot \sum _{k=1}^{n}k-n}
∑
k
=
1
n
(
2
⋅
k
−
1
)
=
2
⋅
n
⋅
(
n
+
1
)
2
−
n
{\displaystyle \sum _{k=1}^{n}(2\cdot k-1)=2\cdot {\frac {n\cdot (n+1)}{2}}-n}
∑
k
=
1
n
(
2
⋅
k
−
1
)
=
n
2
+
n
−
n
{\displaystyle \sum _{k=1}^{n}(2\cdot k-1)=n^{2}+n-n}
∑
k
=
1
n
(
2
⋅
k
−
1
)
=
n
2
{\displaystyle \sum _{k=1}^{n}(2\cdot k-1)=n^{2}}
Exemplo 4 Suprimindo um dos elementos do conjunto {
1
,
2
,
.
.
.
,
n
{\displaystyle 1,2,...,n}
16,1. Determine o valor de n e qual foi o elemento suprimido do conjunto para o cálculo da média.
Resolução
∑
k
=
1
n
k
=
1
+
2
+
.
.
.
+
n
=
n
⋅
(
n
+
1
)
2
{\displaystyle \sum _{k=1}^{n}k=1+2+...+n={\frac {n\cdot (n+1)}{2}}}
média aritmética:
n
⋅
(
n
+
1
)
2
n
=
n
⋅
(
n
+
1
)
2
n
=
1
n
∑
k
=
1
n
k
{\displaystyle {\frac {\frac {n\cdot (n+1)}{2}}{n}}={\frac {n\cdot (n+1)}{2n}}={\frac {1}{n}}\sum _{k=1}^{n}k}
média aritmética de
n
−
1
:
1
n
−
1
∑
k
=
1
n
−
1
k
=
16
,
1
{\displaystyle n-1{\text{: }}{\frac {1}{n-1}}\sum _{k=1}^{n-1}k=16,1}
Exemplo 5 Encontre uma fórmula fechada
∑
k
=
1
n
k
3
{\displaystyle \sum _{k=1}^{n}k^{3}}
onde n ∈ N, com n ≥ 1. Observe o padrão utilizado para resolver as duas questões anteriores e
Resolução Exemplo 6 Calcule a soma
∑
k
=
1
n
k
⋅
k
!
{\displaystyle \sum _{k=1}^{n}k\cdot k!}
onde
n
∈
N
,com
n
≥
1
{\displaystyle n\in N{\text{,com }}n\geq 1}
Resolução Separando o somatório:
∑
k
=
1
n
k
⋅
k
!
=
∑
k
=
1
n
k
⋅
∑
k
=
1
n
k
!
{\displaystyle \sum _{k=1}^{n}k\cdot k!=\sum _{k=1}^{n}k\cdot \sum _{k=1}^{n}k!}
Temos:
∑
k
=
1
n
k
=
n
(
n
+
1
)
2
{\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}}}
e teremos que descobrir o
∑
k
=
1
n
k
!
{\displaystyle \sum _{k=1}^{n}k!}
então
∑
k
=
1
n
k
!
+
(
n
+
1
)
!
=
∑
k
=
0
n
(
k
!
+
1
)
!
{\displaystyle \sum _{k=1}^{n}k!+(n+1)!=\sum _{k=0}^{n}(k!+1)!}
1
+
∑
k
=
1
n
(
k
+
1
)
!
=
1
+
∑
k
=
1
n
(
k
+
1
)
k
!
{\displaystyle 1+\sum _{k=1}^{n}(k+1)!=1+\sum _{k=1}^{n}(k+1)k!}
Exemplo 7 Os números
2
,
3
e
5
{\displaystyle {\sqrt {2}},\quad {\sqrt {3}}\quad {\text{e}}\quad {\sqrt {5}}}
podem pertencer a uma mesma progressão aritmética?
Resolução Assumindo uma PA
(
1
,
2
,
3
,
4
,
5
)
{\displaystyle ({\sqrt {1}},{\sqrt {2}},{\sqrt {3}},{\sqrt {4}},{\sqrt {5}})}
os termos
2
,
3
e
5
{\displaystyle {\sqrt {2}},\quad {\sqrt {3}}\quad {\text{e}}\quad {\sqrt {5}}}
a
+
c
2
=
b
{\displaystyle {\frac {a+c}{2}}=b}
3
=
1
,
732050807568877
{\displaystyle {\sqrt {3}}=1,732050807568877}
1
+
5
2
=
1
,
618033988749895
{\displaystyle {\frac {{\sqrt {1}}+{\sqrt {5}}}{2}}=1,618033988749895}
2
+
4
2
=
1
,
707106781186548
{\displaystyle {\frac {{\sqrt {2}}+{\sqrt {4}}}{2}}=1,707106781186548}
Portanto
2
,
3
e
5
{\displaystyle {\sqrt {2}},\quad {\sqrt {3}}\quad {\text{e}}\quad {\sqrt {5}}}
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)}
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)}
Passo indutivo: s < t Suponha que para um
k
∈
N
,
k
>
s
{\displaystyle k\in N,k>s}
∑
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)}
k
+
1
{\displaystyle k+1}
∑
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)}
∑
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)}
k
−
s
{\displaystyle k-s}
∑
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))}
C
{\displaystyle C}
∑
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)}
∑
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)}
∀
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)}
Passo indutivo: s < t Suponha que para um
k
∈
N
,
k
>
s
{\displaystyle k\in N,k>s}
∑
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)}
k
+
1
{\displaystyle k+1}
∑
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)}
∑
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)}
k
−
s
{\displaystyle k-s}
∑
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)}
k
+
2
{\displaystyle k+2}
∑
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}
![{\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)
