هذه قائمة بالمتسلسلات الرياضياتية والتي تحتوي على صيغ بالتجميعات المنتهية واللامنتهية. ويمكن استخدامها مع غيرها من الأدوات التي تقوم بتقدير التجميعات evaluating sums.
المعادلة
∑
i
=
1
n
i
=
n
(
n
+
1
)
2
{\displaystyle \sum _{i=1}^{n}i={\frac {n(n+1)}{2}}\,\!}
∑
i
=
1
n
i
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
=
n
3
3
+
n
2
2
+
n
6
{\displaystyle \sum _{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}\,\!}
∑
i
=
1
n
i
3
=
(
n
(
n
+
1
)
2
)
2
=
n
4
4
+
n
3
2
+
n
2
4
=
[
∑
i
=
1
n
i
]
2
{\displaystyle \sum _{i=1}^{n}i^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}=\left[\sum _{i=1}^{n}i\right]^{2}\,\!}
∑
i
=
1
n
i
4
=
n
(
n
+
1
)
(
2
n
+
1
)
(
3
n
2
+
3
n
−
1
)
30
{\displaystyle \sum _{i=1}^{n}i^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}\,\!}
∑
i
=
0
n
i
s
=
(
n
+
1
)
s
+
1
s
+
1
+
∑
k
=
1
s
B
k
s
−
k
+
1
(
s
k
)
(
n
+
1
)
s
−
k
+
1
{\displaystyle \sum _{i=0}^{n}i^{s}={\frac {(n+1)^{s+1}}{s+1}}+\sum _{k=1}^{s}{\frac {B_{k}}{s-k+1}}{s \choose k}(n+1)^{s-k+1}\,\!}
حيث أن
B
k
{\displaystyle B_{k}}
عدد بيرنولي ذو العدد k .
∑
i
=
1
∞
i
−
s
=
∏
p
prime
1
1
−
p
−
s
=
ζ
(
s
)
{\displaystyle \sum _{i=1}^{\infty }i^{-s}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}=\zeta (s)\,\!}
حيث أن
ζ
(
s
)
{\displaystyle \zeta (s)}
دالة زيتا .
تجميع اللانهائيات (عندما يكون
|
x
|
<
1
{\displaystyle |x|<1}
تجميع النهائيات
∑
i
=
0
∞
x
i
=
1
1
−
x
{\displaystyle \sum _{i=0}^{\infty }x^{i}={\frac {1}{1-x}}\,\!}
∑
i
=
0
n
x
i
=
1
−
x
n
+
1
1
−
x
=
1
+
1
r
(
1
−
1
(
1
+
r
)
n
)
,
w
h
e
r
e
r
>
0
a
n
d
x
=
1
1
+
r
.
{\displaystyle \sum _{i=0}^{n}x^{i}={\frac {1-x^{n+1}}{1-x}}=1+{\frac {1}{r}}(1-{\frac {1}{(1+r)^{n}}}),\mathrm {where} \ r>0\ \mathrm {and} \ x={\frac {1}{1+r}}.\,\!}
∑
i
=
0
∞
x
2
i
=
1
1
−
x
2
{\displaystyle \sum _{i=0}^{\infty }x^{2i}={\frac {1}{1-x^{2}}}\,\!}
∑
i
=
1
∞
i
x
i
=
x
(
1
−
x
)
2
{\displaystyle \sum _{i=1}^{\infty }ix^{i}={\frac {x}{(1-x)^{2}}}\,\!}
∑
i
=
1
n
i
x
i
=
x
1
−
x
n
(
1
−
x
)
2
−
n
x
n
+
1
1
−
x
{\displaystyle \sum _{i=1}^{n}ix^{i}=x{\frac {1-x^{n}}{(1-x)^{2}}}-{\frac {nx^{n+1}}{1-x}}\,\!}
∑
i
=
1
∞
i
2
x
i
=
x
(
1
+
x
)
(
1
−
x
)
3
{\displaystyle \sum _{i=1}^{\infty }i^{2}x^{i}={\frac {x(1+x)}{(1-x)^{3}}}\,\!}
∑
i
=
1
n
i
2
x
i
=
x
(
1
+
x
−
(
n
+
1
)
2
x
n
+
(
2
n
2
+
2
n
−
1
)
x
n
+
1
−
n
2
x
n
+
2
)
(
1
−
x
)
3
{\displaystyle \sum _{i=1}^{n}i^{2}x^{i}={\frac {x(1+x-(n+1)^{2}x^{n}+(2n^{2}+2n-1)x^{n+1}-n^{2}x^{n+2})}{(1-x)^{3}}}\,\!}
∑
i
=
1
∞
i
3
x
i
=
x
(
1
+
4
x
+
x
2
)
(
1
−
x
)
4
{\displaystyle \sum _{i=1}^{\infty }i^{3}x^{i}={\frac {x(1+4x+x^{2})}{(1-x)^{4}}}\,\!}
∑
i
=
1
∞
i
4
x
i
=
x
(
1
+
x
)
(
1
+
10
x
+
x
2
)
(
1
−
x
)
5
{\displaystyle \sum _{i=1}^{\infty }i^{4}x^{i}={\frac {x(1+x)(1+10x+x^{2})}{(1-x)^{5}}}\,\!}
∑
i
=
1
∞
i
k
x
i
=
Li
−
k
(
x
)
,
{\displaystyle \sum _{i=1}^{\infty }i^{k}x^{i}=\operatorname {Li} _{-k}(x),\,\!}
s x ) هو لوغاريتم متعدد للمتغير x .
المعادلة
∑
i
=
1
∞
x
i
i
=
−
ln
(
1
−
x
)
for
|
x
|
≤
1
,
x
≠
1
{\displaystyle \sum _{i=1}^{\infty }{\frac {x^{i}}{i}}=-\ln \left({1-x}\right)\quad {\mbox{ for }}|x|\leq 1,\,x\not =1\,\!}
∑
i
=
0
∞
(
−
1
)
i
2
i
+
1
x
2
i
+
1
=
x
−
x
3
3
+
x
5
5
−
⋯
=
arctan
(
x
)
{\displaystyle \sum _{i=0}^{\infty }{\frac {(-1)^{i}}{2i+1}}x^{2i+1}=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots =\arctan(x)\,\!}
∑
i
=
0
∞
x
2
i
+
1
2
i
+
1
=
a
r
c
t
a
n
h
(
x
)
for
|
x
|
<
1
{\displaystyle \sum _{i=0}^{\infty }{\frac {x^{2i+1}}{2i+1}}=\mathrm {arctanh} (x)\quad {\mbox{ for }}|x|<1\,\!}
∑
i
=
1
∞
1
i
2
=
π
2
6
{\displaystyle \sum _{i=1}^{\infty }{\frac {1}{i^{2}}}={\frac {\pi ^{2}}{6}}\,\!}
هي متسلسلة متعددة القوى نشأت من مبرهنة تايلور ويكون لديها معامل عاملي .
المعادلة
∑
i
=
0
∞
x
i
i
!
=
e
x
{\displaystyle \sum _{i=0}^{\infty }{\frac {x^{i}}{i!}}=e^{x}}
∑
i
=
0
∞
i
x
i
i
!
=
x
e
x
{\displaystyle \sum _{i=0}^{\infty }i{\frac {x^{i}}{i!}}=xe^{x}}
توزيع بواسون )
∑
i
=
0
∞
i
2
x
i
i
!
=
(
x
+
x
2
)
e
x
{\displaystyle \sum _{i=0}^{\infty }i^{2}{\frac {x^{i}}{i!}}=(x+x^{2})e^{x}}
العزم الثاني لتوزيع بواسون)
∑
i
=
0
∞
i
3
x
i
i
!
=
(
x
+
3
x
2
+
x
3
)
e
x
{\displaystyle \sum _{i=0}^{\infty }i^{3}{\frac {x^{i}}{i!}}=(x+3x^{2}+x^{3})e^{x}}
∑
i
=
0
∞
i
4
x
i
i
!
=
(
x
+
7
x
2
+
6
x
3
+
x
4
)
e
x
{\displaystyle \sum _{i=0}^{\infty }i^{4}{\frac {x^{i}}{i!}}=(x+7x^{2}+6x^{3}+x^{4})e^{x}}
∑
i
=
0
∞
(
−
1
)
i
(
2
i
+
1
)
!
x
2
i
+
1
=
x
−
x
3
3
!
+
x
5
5
!
−
⋯
=
sin
x
{\displaystyle \sum _{i=0}^{\infty }{\frac {(-1)^{i}}{(2i+1)!}}x^{2i+1}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots =\sin x}
∑
i
=
0
∞
(
−
1
)
i
(
2
i
)
!
x
2
i
=
1
−
x
2
2
!
+
x
4
4
!
−
⋯
=
cos
x
{\displaystyle \sum _{i=0}^{\infty }{\frac {(-1)^{i}}{(2i)!}}x^{2i}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots =\cos x}
∑
i
=
0
∞
x
2
i
+
1
(
2
i
+
1
)
!
=
sinh
x
{\displaystyle \sum _{i=0}^{\infty }{\frac {x^{2i+1}}{(2i+1)!}}=\sinh x}
∑
i
=
0
∞
x
2
i
(
2
i
)
!
=
cosh
x
{\displaystyle \sum _{i=0}^{\infty }{\frac {x^{2i}}{(2i)!}}=\cosh x}
عدل
المعادلة
∑
n
=
0
∞
(
2
n
)
!
4
n
(
n
!
)
2
(
2
n
+
1
)
x
2
n
+
1
=
arcsin
x
for
|
x
|
<
1
{\displaystyle \sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}=\arcsin x\quad {\mbox{ for }}|x|<1\!}
∑
i
=
0
∞
(
−
1
)
i
(
2
i
)
!
4
i
(
i
!
)
2
(
2
i
+
1
)
x
2
i
+
1
=
a
r
c
s
i
n
h
(
x
)
for
|
x
|
<
1
{\displaystyle \sum _{i=0}^{\infty }{\frac {(-1)^{i}(2i)!}{4^{i}(i!)^{2}(2i+1)}}x^{2i+1}=\mathrm {arcsinh} (x)\quad {\mbox{ for }}|x|<1\!}
متسلسلة ثنائية الحد (و من ضمنها متسلسلة الجذر التربيعي عندما يكون
α
=
1
/
2
{\displaystyle \alpha =1/2}
α
=
−
1
{\displaystyle \alpha =-1}
الجذر التربيعي :
1
+
x
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
)
!
(
1
−
2
n
)
n
!
2
4
n
x
n
for
|
x
|
<
1
{\displaystyle {\sqrt {1+x}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{(1-2n)n!^{2}4^{n}}}x^{n}\quad {\mbox{ for }}|x|<1\!}
المتسلسلة الهندسية :
(
1
+
x
)
−
1
=
∑
n
=
0
∞
(
−
1
)
n
x
n
for
|
x
|
<
1
{\displaystyle (1+x)^{-1}=\sum _{n=0}^{\infty }(-1)^{n}x^{n}\quad {\mbox{ for }}|x|<1}
الصيغة العامة:
(
1
+
x
)
α
=
∑
n
=
0
∞
(
α
n
)
x
n
for all
|
x
|
<
1
and all complex
α
{\displaystyle (1+x)^{\alpha }=\sum _{n=0}^{\infty }{\alpha \choose n}x^{n}\quad {\mbox{ for all }}|x|<1{\mbox{ and all complex }}\alpha \!}
مع العوامل الثنائية الحد المعممة
(
α
n
)
=
∏
k
=
1
n
α
−
k
+
1
k
=
α
(
α
−
1
)
⋯
(
α
−
n
+
1
)
n
!
{\displaystyle {\alpha \choose n}=\prod _{k=1}^{n}{\frac {\alpha -k+1}{k}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}\!}
[ 1]
∑
i
=
0
∞
(
i
+
n
i
)
x
i
=
1
(
1
−
x
)
n
+
1
{\displaystyle \sum _{i=0}^{\infty }{i+n \choose i}x^{i}={\frac {1}{(1-x)^{n+1}}}}
[ 1]
∑
i
=
0
∞
1
i
+
1
(
2
i
i
)
x
i
=
1
2
x
(
1
−
4
x
)
{\displaystyle \sum _{i=0}^{\infty }{\frac {1}{i+1}}{2i \choose i}x^{i}={\frac {1}{2x}}({\sqrt {1-4x}})}
[ 1]
∑
i
=
0
∞
(
2
i
i
)
x
i
=
1
1
−
4
x
{\displaystyle \sum _{i=0}^{\infty }{2i \choose i}x^{i}={\frac {1}{\sqrt {1-4x}}}}
[ 1]
∑
i
=
0
∞
(
2
i
+
n
i
)
x
i
=
1
1
−
4
x
(
1
−
1
−
4
x
2
x
)
n
{\displaystyle \sum _{i=0}^{\infty }{2i+n \choose i}x^{i}={\frac {1}{\sqrt {1-4x}}}\left({\frac {1-{\sqrt {1-4x}}}{2x}}\right)^{n}}
المعادلة
∑
i
=
0
n
(
n
i
)
=
2
n
{\displaystyle \sum _{i=0}^{n}{n \choose i}=2^{n}}
∑
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}}
∑
i
=
0
n
(
−
1
)
i
(
n
i
)
=
0
{\displaystyle \sum _{i=0}^{n}(-1)^{i}{n \choose i}=0}
∑
i
=
0
n
(
i
k
)
=
(
n
+
1
k
+
1
)
{\displaystyle \sum _{i=0}^{n}{i \choose k}={n+1 \choose k+1}}
∑
i
=
0
n
(
k
+
i
i
)
=
(
k
+
n
+
1
n
)
{\displaystyle \sum _{i=0}^{n}{k+i \choose i}={k+n+1 \choose n}}
∑
i
=
0
r
(
r
i
)
(
s
n
−
i
)
=
(
r
+
s
n
)
{\displaystyle \sum _{i=0}^{r}{r \choose i}{s \choose n-i}={r+s \choose n}}
إن تجميعات الجيوب والجيوب التمام مأخوذة من متسلسلة فوييه .
المعادلة
∑
i
=
1
n
sin
(
i
π
n
)
=
0
{\displaystyle \sum _{i=1}^{n}\sin \left({\frac {i\pi }{n}}\right)=0}
∑
i
=
1
n
cos
(
i
π
n
)
=
0
{\displaystyle \sum _{i=1}^{n}\cos \left({\frac {i\pi }{n}}\right)=0}
المعادلة
∑
n
=
b
+
1
∞
b
n
2
−
b
2
=
∑
n
=
1
2
b
1
2
n
{\displaystyle \sum _{n=b+1}^{\infty }{\frac {b}{n^{2}-b^{2}}}=\sum _{n=1}^{2b}{\frac {1}{2n}}}
![{\displaystyle \sum _{i=1}^{n}i^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}=\left[\sum _{i=1}^{n}i\right]^{2}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ace1e8ca0a0e9d694f2181b6c21b377041abbb9f)
