In mathematics, summation is the addition of a sequence of numbers. The result is a sum or total . The sum of a sequence of numbers is denoted with an enlarged capital Greek sigma symbol
∑
{\displaystyle \textstyle \sum }
Terminology [ edit ] The summation symbol The sum of a sequence of numbers is denoted by:
∑
i
=
1
n
a
i
=
a
1
+
a
2
+
a
3
+
⋯
+
a
n
−
1
+
a
n
{\displaystyle \sum _{i\mathop {=} 1}^{n}a_{i}=a_{1}+a_{2}+a_{3}+\cdots +a_{n-1}+a_{n}}
where i represents the index ; ai are the successive terms in the sum; 1 is the lower bound , and n is the upper bound . The index, i , is incremented by 1 for each successive term, stopping when i = n The numbers to be summed are called addends , or sometimes summands The addends, represented by ai , may be integers, rational numbers , real numbers , or complex numbers . .[1]
Formulae [ edit ]
∑
i
=
1
n
1
=
n
{\displaystyle \sum _{i=1}^{n}1=n\quad }
∑
i
=
1
n
c
=
n
c
{\displaystyle \sum _{i=1}^{n}c=nc\quad }
c
∑
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}}\qquad }
arithmetic progression , consisting of the n first natural numbers .)[2]
∑
i
=
1
n
2
i
−
1
=
n
2
{\displaystyle \sum _{i=1}^{n}2i-1=n^{2}\qquad }
∑
i
=
0
n
2
i
=
n
(
n
+
1
)
{\displaystyle \sum _{i=0}^{n}2i=n(n+1)\qquad }
∑
i
=
1
n
log
i
=
log
n
!
{\displaystyle \sum _{i=1}^{n}\log i=\log n!\qquad }
logarithms is the logarithm of the product)
∑
i
=
0
n
i
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
=
n
3
3
+
n
2
2
+
n
6
{\displaystyle \sum _{i=0}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}\qquad }
squares , see square pyramidal number .) [2]
∑
i
=
0
n
i
3
=
(
∑
i
=
0
n
i
)
2
=
(
n
(
n
+
1
)
2
)
2
=
n
4
4
+
n
3
2
+
n
2
4
{\displaystyle \sum _{i=0}^{n}i^{3}=\left(\sum _{i=0}^{n}i\right)^{2}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}\qquad }
Nicomachus's theorem ) [2]
∑
i
=
0
n
i
4
=
n
(
n
+
1
)
(
2
n
+
1
)
(
3
n
2
+
3
n
−
1
)
30
=
n
5
5
+
n
4
2
+
n
3
3
−
n
30
{\displaystyle \sum _{i=0}^{n}i^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}={\frac {n^{5}}{5}}+{\frac {n^{4}}{2}}+{\frac {n^{3}}{3}}-{\frac {n}{30}}}
[2]
∑
i
=
1
n
i
5
=
n
2
(
n
+
1
)
2
(
2
n
2
+
2
n
−
1
)
12
{\displaystyle \sum _{i=1}^{n}i^{5}={\frac {n^{2}(n+1)^{2}(2n^{2}+2n-1)}{12}}}
[2]
∑
i
=
1
n
i
6
=
n
(
n
+
1
)
(
2
n
+
1
)
(
3
n
4
+
6
n
3
−
3
n
+
1
)
42
{\displaystyle \sum _{i=1}^{n}i^{6}={\frac {n(n+1)(2n+1)(3n^{4}+6n^{3}-3n+1)}{42}}}
[2]
∑
i
=
1
n
i
7
=
n
2
(
n
+
1
)
2
(
3
n
4
+
6
n
3
−
n
2
−
4
n
+
2
)
24
{\displaystyle \sum _{i=1}^{n}i^{7}={\frac {n^{2}(n+1)^{2}(3n^{4}+6n^{3}-n^{2}-4n+2)}{24}}}
[2]
∑
i
=
1
n
i
8
=
n
(
n
+
1
)
(
2
n
+
1
)
(
5
n
6
+
15
n
5
+
5
n
4
−
15
n
3
−
n
2
−
9
n
−
3
)
90
{\displaystyle \sum _{i=1}^{n}i^{8}={\frac {n(n+1)(2n+1)(5n^{6}+15n^{5}+5n^{4}-15n^{3}-n^{2}-9n-3)}{90}}}
[2]
∑
i
=
1
n
i
9
=
n
2
(
n
+
1
)
2
(
2
n
6
+
6
n
5
+
n
4
−
8
n
3
+
n
2
+
6
n
−
3
)
20
{\displaystyle \sum _{i=1}^{n}i^{9}={\frac {n^{2}(n+1)^{2}(2n^{6}+6n^{5}+n^{4}-8n^{3}+n^{2}+6n-3)}{20}}}
[2]
∑
i
=
1
n
i
10
=
n
(
n
+
1
)
(
2
n
+
1
)
(
3
n
8
+
12
n
7
+
8
n
6
−
18
n
5
−
10
n
4
+
24
n
3
+
2
n
2
−
15
n
+
5
)
66
{\displaystyle \sum _{i=1}^{n}i^{10}={\frac {n(n+1)(2n+1)(3n^{8}+12n^{7}+8n^{6}-18n^{5}-10n^{4}+24n^{3}+2n^{2}-15n+5)}{66}}}
[2]
∑
i
=
1
n
3
i
2
−
3
i
+
1
=
n
3
{\displaystyle \sum _{i=1}^{n}3i^{2}-3i+1=n^{3}}
∑
i
=
1
n
4
i
3
−
6
i
2
+
4
i
−
1
=
n
4
{\displaystyle \sum _{i=1}^{n}4i^{3}-6i^{2}+4i-1=n^{4}}
∑
i
=
1
n
5
i
4
−
10
i
3
+
10
i
2
−
5
i
+
1
=
n
5
{\displaystyle \sum _{i=1}^{n}5i^{4}-10i^{3}+10i^{2}-5i+1=n^{5}}
∑
i
=
1
n
6
i
5
−
15
i
4
+
20
i
3
−
15
i
2
+
6
i
−
1
=
n
6
{\displaystyle \sum _{i=1}^{n}6i^{5}-15i^{4}+20i^{3}-15i^{2}+6i-1=n^{6}}
∑
i
=
1
n
7
i
6
−
21
i
5
+
35
i
4
−
35
i
3
+
21
i
2
−
7
i
+
1
=
n
7
{\displaystyle \sum _{i=1}^{n}7i^{6}-21i^{5}+35i^{4}-35i^{3}+21i^{2}-7i+1=n^{7}}
∑
i
=
1
n
8
i
7
−
28
i
6
+
56
i
5
−
70
i
4
+
56
i
3
−
28
i
2
+
8
i
−
1
=
n
8
{\displaystyle \sum _{i=1}^{n}8i^{7}-28i^{6}+56i^{5}-70i^{4}+56i^{3}-28i^{2}+8i-1=n^{8}}
∑
i
=
1
n
9
i
8
−
36
i
7
+
84
i
6
−
126
i
5
+
126
i
4
−
84
i
3
+
36
i
2
−
9
i
+
1
=
n
9
{\displaystyle \sum _{i=1}^{n}9i^{8}-36i^{7}+84i^{6}-126i^{5}+126i^{4}-84i^{3}+36i^{2}-9i+1=n^{9}}
∑
i
=
1
n
10
i
9
−
45
i
8
+
120
i
7
−
210
i
6
+
252
i
5
−
210
i
4
+
120
i
3
−
45
i
2
+
10
i
−
1
=
n
10
{\displaystyle \sum _{i=1}^{n}10i^{9}-45i^{8}+120i^{7}-210i^{6}+252i^{5}-210i^{4}+120i^{3}-45i^{2}+10i-1=n^{10}}
∑
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}}}
a
≠
1
{\displaystyle a\neq 1}
geometric series )
∑
i
=
0
n
−
1
1
2
i
=
2
−
1
2
n
−
1
{\displaystyle \sum _{i=0}^{n-1}{\frac {1}{2^{i}}}=2-{\frac {1}{2^{n-1}}}}
∑
i
=
0
n
−
1
i
a
i
=
a
−
n
a
n
+
(
n
−
1
)
a
n
+
1
(
1
−
a
)
2
{\displaystyle \sum _{i=0}^{n-1}ia^{i}={\frac {a-na^{n}+(n-1)a^{n+1}}{(1-a)^{2}}}}
a
≠
1
{\displaystyle a\neq 1}
∑
i
=
0
n
−
1
i
2
i
=
2
+
(
n
−
2
)
2
n
{\displaystyle \sum _{i=0}^{n-1}i2^{i}=2+(n-2)2^{n}}
∑
i
=
0
n
−
1
i
2
i
=
2
−
n
+
1
2
n
−
1
{\displaystyle \sum _{i=0}^{n-1}{\frac {i}{2^{i}}}=2-{\frac {n+1}{2^{n-1}}}}
∑
i
=
0
n
−
1
(
b
+
i
d
)
a
i
=
b
−
[
b
+
(
n
−
1
)
d
]
a
n
1
−
a
+
d
a
(
1
−
a
n
−
1
)
(
1
−
a
)
2
{\displaystyle {\begin{aligned}\sum _{i=0}^{n-1}\left(b+id\right)a^{i}&={\frac {b-[b+(n-1)d]a^{n}}{1-a}}+{\frac {da(1-a^{n-1})}{(1-a)^{2}}}\end{aligned}}}
a
≠
1
{\displaystyle a\neq 1}
arithmetico-geometric series )
∑
i
=
0
n
(
n
i
)
=
2
n
{\displaystyle \sum _{i=0}^{n}{n \choose i}=2^{n}}
∑
i
=
0
n
(
n
i
)
p
i
(
1
−
p
)
n
−
i
=
1
{\displaystyle \sum _{i=0}^{n}{n \choose i}p^{i}(1-p)^{n-i}=1}
0
≤
p
≤
1.
{\displaystyle 0\leq p\leq 1.}
∑
k
=
0
m
(
n
+
k
n
)
=
(
n
+
m
+
1
n
+
1
)
{\displaystyle \sum _{k=0}^{m}\left({\begin{array}{c}n+k\\n\\\end{array}}\right)=\left({\begin{array}{c}n+m+1\\n+1\\\end{array}}\right)}
∑
i
=
1
n
i
(
n
i
)
=
n
(
2
n
−
1
)
{\displaystyle \sum _{i=1}^{n}i{n \choose i}=n(2^{n-1})}
∑
i
=
0
n
(
n
i
)
i
+
1
=
2
n
+
1
−
1
n
+
1
{\displaystyle \sum _{i=0}^{n}{\frac {n \choose i}{i+1}}={\frac {2^{n+1}-1}{n+1}}}
∑
i
=
k
n
(
i
k
)
=
(
n
+
1
k
+
1
)
{\displaystyle \sum _{i=k}^{n}{i \choose k}={n+1 \choose k+1}}
∑
i
=
0
n
(
n
i
)
a
n
−
i
b
i
=
(
a
+
b
)
n
{\displaystyle \sum _{i=0}^{n}{n \choose i}a^{n-i}b^{i}=(a+b)^{n}}
binomial theorem
∑
i
=
0
n
i
⋅
i
!
=
(
n
+
1
)
!
−
1
{\displaystyle \sum _{i=0}^{n}i\cdot i!=(n+1)!-1}
∑
i
=
0
n
(
m
+
i
−
1
i
)
=
(
m
+
n
n
)
{\displaystyle \sum _{i=0}^{n}{m+i-1 \choose i}={m+n \choose n}}
∑
i
=
0
n
(
n
i
)
2
=
(
2
n
n
)
{\displaystyle \sum _{i=0}^{n}{n \choose i}^{2}={2n \choose n}}
See also [ edit ]
↑ Graham, Ronald L.; Knuth, Donald E.; Patashnik, Oren (1994). "Chapter 2: Sums". Concrete Mathematics: A Foundation for Computer Science (2nd Edition) . Addison-Wesley Professional. CS1 maint: Uses authors parameter (link ) Search this book on
↑ 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 W. H. Boyer (Editor), CRC Standard Math Tables, CRC Press, p 52, 1984
External links [ edit ] References [ edit ]
This article "Summation Formula List" is from Wikipedia . The list of its authors can be seen in its historical and/or the page Edithistory:Summation Formula List . Articles copied from Draft Namespace on Wikipedia could be seen on the Draft Namespace of Wikipedia and not main one.
![{\displaystyle {\begin{aligned}\sum _{i=0}^{n-1}\left(b+id\right)a^{i}&={\frac {b-[b+(n-1)d]a^{n}}{1-a}}+{\frac {da(1-a^{n-1})}{(1-a)^{2}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2df3abea8017bb5e181961e2bccda46fefabab5)
