suma częściowa
Suma częściowa ciągu to inaczej suma kilku kolejnych wyrazów pewnego ciągu. Najprostszym przykładem może być
a
1
+
a
2
{\displaystyle a_{1}+a_{2}}
a
2
+
a
4
+
a
6
{\displaystyle a_{2}+a_{4}+a_{6}}
(
a
n
)
{\displaystyle (a_{n})}
Policzmy sumę czterech kolejnych wyrazów ciągu
(
a
n
)
{\displaystyle (a_{n})}
a
n
=
2
⋅
|
n
−
3
|
{\displaystyle a_{n}=2\cdot |n-3|}
a
1
=
2
⋅
|
1
−
3
|
=
2
⋅
2
=
4
{\displaystyle a_{1}=2\cdot |1-3|=2\cdot 2=4}
a
2
=
2
⋅
|
2
−
3
|
=
2
{\displaystyle a_{2}=2\cdot |2-3|=2}
a
3
=
2
⋅
|
3
−
3
|
=
0
{\displaystyle a_{3}=2\cdot |3-3|=0}
a
4
=
2
⋅
|
4
−
3
|
=
2
{\displaystyle a_{4}=2\cdot |4-3|=2}
a
1
+
a
2
+
a
3
+
a
4
=
4
+
2
+
0
+
2
=
8
{\displaystyle a_{1}+a_{2}+a_{3}+a_{4}=4+2+0+2=8}
Podobnie policzmy sumę wyrazów
c
2
+
c
10
+
c
30
+
c
51
+
c
1001
{\displaystyle c_{2}+c_{10}+c_{30}+c_{51}+c_{1001}}
(
c
n
)
{\displaystyle (c_{n})}
c
1
=
10
{\displaystyle c_{1}=10}
-3 . Jednak najpierw musimy policzyć ile wynoszą odpowiednie wyrazy. Ze wzoru na n-ty wyraz ciągu arytmetycznego mamy:
c
2
=
10
+
(
2
−
1
)
⋅
(
−
3
)
=
7
{\displaystyle c_{2}=10+(2-1)\cdot (-3)=7}
c
10
=
10
+
(
10
−
1
)
⋅
(
−
3
)
=
−
17
{\displaystyle c_{10}=10+(10-1)\cdot (-3)=-17}
c
30
=
10
+
29
⋅
(
−
3
)
=
−
77
{\displaystyle c_{30}=10+29\cdot (-3)=-77}
c
51
=
10
+
50
⋅
(
−
3
)
=
−
140
{\displaystyle c_{51}=10+50\cdot (-3)=-140}
c
1001
=
10
+
1000
⋅
(
−
3
)
=
−
2990
{\displaystyle c_{1001}=10+1000\cdot (-3)=-2990}
Zatem suma
c
2
+
c
10
+
c
30
+
c
51
+
c
1001
=
7
−
17
−
77
−
140
−
2990
=
−
3217
{\displaystyle c_{2}+c_{10}+c_{30}+c_{51}+c_{1001}=7-17-77-140-2990=-3217}
Sumę kolejnych n wyrazów pewnego ciągu, czyli
a
1
+
a
2
+
a
3
+
a
4
+
⋯
+
a
n
{\displaystyle a_{1}+a_{2}+a_{3}+a_{4}+\dots +a_{n}}
S
n
{\displaystyle S_{n}}
S
5
=
a
1
+
a
2
+
a
3
+
a
4
+
a
5
{\displaystyle S_{5}=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}}
S
3
=
a
1
+
a
2
+
a
3
{\displaystyle S_{3}=a_{1}+a_{2}+a_{3}}
S
50
=
a
1
+
a
2
+
a
3
+
a
4
+
⋯
+
a
50
{\displaystyle S_{50}=a_{1}+a_{2}+a_{3}+a_{4}+\dots +a_{50}}
S
1
=
a
1
{\displaystyle S_{1}=a_{1}}
Używając tego oznaczenia możemy zapisać także sumę kolejnych, ale nie koniecznie początkowych wyrazów, na przykład:
a
3
+
a
4
+
a
5
=
(
a
1
+
a
2
+
a
3
+
a
4
+
a
5
)
−
(
a
1
+
a
2
)
=
S
5
−
S
2
{\displaystyle a_{3}+a_{4}+a_{5}=(a_{1}+a_{2}+a_{3}+a_{4}+a_{5})-(a_{1}+a_{2})=S_{5}-S_{2}}
a
5
+
a
6
+
a
7
=
(
a
1
+
a
2
+
a
3
+
a
4
+
a
5
+
a
6
+
a
7
)
−
(
a
1
+
a
2
+
a
3
+
a
4
)
=
S
7
−
S
4
{\displaystyle a_{5}+a_{6}+a_{7}=(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7})-(a_{1}+a_{2}+a_{3}+a_{4})=S_{7}-S_{4}}
a
50
+
a
51
+
a
52
+
⋯
+
a
100
=
(
a
1
+
a
2
+
⋯
+
a
49
+
a
50
+
⋯
+
a
100
)
−
(
a
1
+
a
2
+
⋯
+
a
49
)
=
S
100
−
S
49
{\displaystyle a_{50}+a_{51}+a_{52}+\dots +a_{100}=(a_{1}+a_{2}+\dots +a_{49}+a_{50}+\dots +a_{100})-(a_{1}+a_{2}+\dots +a_{49})=S_{100}-S_{49}}
W ogólności suma
a
k
+
a
k
+
1
+
⋯
+
a
n
=
(
a
1
+
a
2
+
⋯
+
a
k
−
1
+
a
k
+
a
n
)
−
(
a
1
+
a
2
+
⋯
+
a
k
−
1
)
=
S
n
−
S
k
−
1
{\displaystyle a_{k}+a_{k+1}+\dots +a_{n}=(a_{1}+a_{2}+\dots +a_{k-1}+a_{k}+a_{n})-(a_{1}+a_{2}+\dots +a_{k-1})=S_{n}-S_{k-1}}
edytuj
suma częściowa ciągu arytmetycznego
TWIERDZENIE Suma n początkowych wyrazów ciągu arytmetycznego wynosi:
S
n
=
a
1
+
a
2
+
a
3
+
⋯
+
a
n
=
a
1
+
a
n
2
⋅
n
{\displaystyle S_{n}=a_{1}+a_{2}+a_{3}+\dots +a_{n}={\frac {a_{1}+a_{n}}{2}}\cdot n}
Znając to twierdzenie możemy policzyć sumę
S
10
=
1
+
2
+
3
+
⋯
+
10
{\displaystyle S_{10}=1+2+3+\dots +10}
n
=
10
{\displaystyle n=10}
a
1
=
1
{\displaystyle a_{1}=1}
a
10
=
10
{\displaystyle a_{10}=10}
S
10
=
a
1
+
a
n
2
⋅
n
=
1
+
10
2
⋅
10
=
55
{\displaystyle S_{10}={\frac {a_{1}+a_{n}}{2}}\cdot n={\frac {1+10}{2}}\cdot 10=55}
Korzystając z tego wzoru możemy w bardzo prosty sposób znaleźć wzór na sumę n kolejnych liczb naturalnych. Zero możemy pominąć, ponieważ nic nie wnosi do naszej sumy. Zobaczmy -- pierwszą liczbą będzie 1 , czyli
a
1
=
1
{\displaystyle a_{1}=1}
a
n
=
n
{\displaystyle a_{n}=n}
S
n
=
1
+
2
+
3
+
⋯
+
n
=
1
+
n
2
⋅
n
=
n
(
n
+
1
)
2
{\displaystyle S_{n}=1+2+3+\dots +n={\frac {1+n}{2}}\cdot n={\frac {n(n+1)}{2}}}
być może już przez niektórych znany.
Policzmy teraz sumę trzydziestu jeden kolejnych wyrazów ciągu
(
t
n
)
{\displaystyle (t_{n})}
t
1
=
10
{\displaystyle t_{1}=10}
r
=
4
{\displaystyle r=4}
n
=
31
{\displaystyle n=31}
t
31
{\displaystyle t_{31}}
t
31
=
10
+
(
31
−
1
)
⋅
4
=
10
+
120
=
130
{\displaystyle t_{31}=10+(31-1)\cdot 4=10+120=130}
Teraz tylko zastosować wzór na sumę początkowych wyrazów:
S
n
=
t
1
+
t
2
+
t
3
+
⋯
+
t
31
=
t
1
+
t
31
2
⋅
31
=
{\displaystyle S_{n}=t_{1}+t_{2}+t_{3}+\dots +t_{31}={\frac {t_{1}+t_{31}}{2}}\cdot 31=}
=
10
+
130
2
⋅
31
=
2170
{\displaystyle ={\frac {10+130}{2}}\cdot 31=2170}
Znajdźmy wzór na sume n początkowych wyrazów ciągu znając jedynie n ,
a
1
{\displaystyle a_{1}}
r . Wiemy ze wzoru na n-ty wyraz, że
a
n
=
a
1
+
(
n
−
1
)
⋅
r
{\displaystyle a_{n}=a_{1}+(n-1)\cdot r}
S
n
=
a
1
+
a
n
2
⋅
n
=
a
1
+
a
1
+
(
n
−
1
)
⋅
r
2
⋅
n
{\displaystyle S_{n}={\frac {a_{1}+a_{n}}{2}}\cdot n={\frac {a_{1}+a_{1}+(n-1)\cdot r}{2}}\cdot n}
Po drobnym przekształceniach mamy:
S
n
=
[
2
a
1
+
(
n
−
1
)
⋅
r
]
⋅
n
2
{\displaystyle S_{n}={\frac {[2a_{1}+(n-1)\cdot r]\cdot n}{2}}}
(suma n początkowych wyrazów ciągu arytmetycznego o różnicy r )
Czy wzór
S
n
=
a
1
+
a
n
2
⋅
n
{\displaystyle S_{n}={\frac {a_{1}+a_{n}}{2}}\cdot n}
Dowód:
Wiemy, że
S
n
=
a
1
+
a
2
+
⋯
+
a
n
{\displaystyle S_{n}=a_{1}+a_{2}+\dots +a_{n}}
(
a
n
)
{\displaystyle (a_{n})}
a
k
=
a
1
+
(
k
−
1
)
⋅
r
{\displaystyle a_{k}=a_{1}+(k-1)\cdot r}
S
n
=
a
1
+
a
2
+
⋯
+
a
n
=
[
a
1
+
(
1
−
1
)
⋅
r
]
+
[
a
1
+
(
2
−
1
)
⋅
r
]
+
⋯
+
[
a
1
+
(
n
−
1
)
⋅
r
]
{\displaystyle S_{n}=a_{1}+a_{2}+\dots +a_{n}=[a_{1}+(1-1)\cdot r]+[a_{1}+(2-1)\cdot r]+\dots +[a_{1}+(n-1)\cdot r]}
sumę tę możemy także przepisać jako (idąc od końca do początku):
S
n
=
[
a
1
+
(
n
−
1
)
⋅
r
]
+
[
a
1
+
(
n
−
2
)
⋅
r
]
+
⋯
+
[
a
1
+
(
2
−
1
)
⋅
r
]
+
[
a
1
+
(
1
−
1
)
⋅
r
]
{\displaystyle S_{n}=[a_{1}+(n-1)\cdot r]+[a_{1}+(n-2)\cdot r]+\dots +[a_{1}+(2-1)\cdot r]+[a_{1}+(1-1)\cdot r]}
Dodając obydwie sumy do siebie otrzymujemy:
S
n
{\displaystyle S_{n}}
=
{\displaystyle =}
[
a
1
+
(
1
−
1
)
⋅
r
]
{\displaystyle [a_{1}+(1-1)\cdot r]}
+
{\displaystyle +}
[
a
1
+
(
2
−
1
)
⋅
r
]
{\displaystyle [a_{1}+(2-1)\cdot r]}
+
{\displaystyle +}
[
a
1
+
(
3
−
1
)
⋅
r
]
{\displaystyle [a_{1}+(3-1)\cdot r]}
+
{\displaystyle +}
…
{\displaystyle \dots }
+
{\displaystyle +}
[
a
1
+
(
n
−
1
)
⋅
r
]
{\displaystyle [a_{1}+(n-1)\cdot r]}
+
{\displaystyle +}
S
n
{\displaystyle S_{n}}
=
{\displaystyle =}
[
a
1
+
(
n
−
1
)
⋅
r
]
{\displaystyle [a_{1}+(n-1)\cdot r]}
+
{\displaystyle +}
[
a
1
+
(
n
−
2
)
⋅
r
]
{\displaystyle [a_{1}+(n-2)\cdot r]}
+
{\displaystyle +}
[
a
1
+
(
n
−
3
)
⋅
r
]
{\displaystyle [a_{1}+(n-3)\cdot r]}
+
{\displaystyle +}
…
{\displaystyle \dots }
+
{\displaystyle +}
[
a
1
+
(
n
−
n
)
⋅
r
]
{\displaystyle [a_{1}+(n-n)\cdot r]}
2
S
n
{\displaystyle 2S_{n}}
=
{\displaystyle =}
[
2
a
1
+
(
n
−
1
)
⋅
r
]
{\displaystyle [2a_{1}+(n-1)\cdot r]}
+
{\displaystyle +}
[
2
a
1
+
(
n
−
1
)
⋅
r
]
{\displaystyle [2a_{1}+(n-1)\cdot r]}
+
{\displaystyle +}
[
2
a
1
+
(
n
−
1
)
⋅
r
]
{\displaystyle [2a_{1}+(n-1)\cdot r]}
+
{\displaystyle +}
…
{\displaystyle \dots }
+
{\displaystyle +}
[
2
a
1
+
(
n
−
1
)
⋅
r
]
{\displaystyle [2a_{1}+(n-1)\cdot r]}
Wszystkie powyższe sumy posiadają n składników, zatem:
2
S
n
=
[
2
a
1
+
(
n
−
1
)
⋅
r
]
⋅
n
{\displaystyle 2S_{n}=[2a_{1}+(n-1)\cdot r]\cdot n}
Po podzieleniu przez dwa mamy:
S
n
=
[
2
a
1
+
(
n
−
1
)
⋅
r
]
⋅
n
2
{\displaystyle S_{n}={\frac {[2a_{1}+(n-1)\cdot r]\cdot n}{2}}}
Czyli dochodzimy do wzoru przedstawionego nieco wyżej.
edytuj
suma częściowa ciągu geometrycznego
Możemy teraz bez problemu obliczyć sumę stu dwójek, czyli
S
100
=
2
+
2
+
2
+
2
+
⋯
+
2
{\displaystyle S_{100}=2+2+2+2+\dots +2}
S
100
=
100
⋅
2
=
200
{\displaystyle S_{100}=100\cdot 2=200}
q
=
1
{\displaystyle q=1}
S
100
=
n
⋅
a
1
=
100
⋅
2
=
200
{\displaystyle S_{100}=n\cdot a_{1}=100\cdot 2=200}
(
b
n
)
{\displaystyle (b_{n})}
b
1
=
11
{\displaystyle b_{1}=11}
b
k
+
1
b
k
=
3
dla
k
∈
Z
+
{\displaystyle {\frac {b_{k+1}}{b_{k}}}=3{\mbox{ dla }}k\in \mathbb {Z} _{+}}
Ponieważ
q
=
3
{\displaystyle q=3}
q
≠
1
{\displaystyle q\neq 1}
S
4
=
a
1
⋅
1
−
q
4
1
−
q
=
11
⋅
1
−
3
4
1
−
3
=
11
⋅
−
80
−
2
=
440
{\displaystyle S_{4}=a_{1}\cdot {\frac {1-q^{4}}{1-q}}=11\cdot {\frac {1-3^{4}}{1-3}}=11\cdot {\frac {-80}{-2}}=440}
S
n
=
1
+
2
+
4
+
8
+
⋯
+
64
{\displaystyle S_{n}=1+2+4+8+\dots +64}
a
1
=
1
{\displaystyle a_{1}=1}
q
=
2
{\displaystyle q=2}
a
n
=
1
⋅
2
n
−
1
{\displaystyle a_{n}=1\cdot 2^{n-1}}
a
n
=
64
{\displaystyle a_{n}=64}
a
n
=
2
n
−
1
=
64
=
2
6
{\displaystyle a_{n}=2^{n-1}=64=2^{6}}
n
−
1
=
6
⟹
n
=
7
{\displaystyle n-1=6\implies n=7}
q
=
2
≠
1
{\displaystyle q=2\neq 1}
S
7
=
a
1
⋅
1
−
q
7
1
−
q
=
1
⋅
1
−
2
7
1
−
2
=
−
127
−
1
=
127
{\displaystyle S_{7}=a_{1}\cdot {\frac {1-q^{7}}{1-q}}=1\cdot {\frac {1-2^{7}}{1-2}}={\frac {-127}{-1}}=127}
(
s
n
)
{\displaystyle (s_{n})}
s
k
=
11
⋅
(
−
10
)
k
−
1
dla
k
∈
Z
+
{\displaystyle s_{k}=11\cdot (-10)^{k-1}{\mbox{ dla }}k\in \mathbb {Z} _{+}}
Pamiętamy, że każdy ciąg geometryczny zdefiniowany jest wzorem:
a
k
=
a
1
⋅
q
k
−
1
dla
k
∈
Z
+
{\displaystyle a_{k}=a_{1}\cdot q^{k-1}{\mbox{ dla }}k\in \mathbb {Z} _{+}}
Zauważmy, że gdybyśmy jako
a
1
{\displaystyle a_{1}}
(
s
n
)
{\displaystyle (s_{n})}
s
1
=
11
{\displaystyle s_{1}=11}
q
=
−
10
{\displaystyle q=-10}
q
≠
0
{\displaystyle q\neq 0}
S
9
=
11
⋅
1
−
(
−
10
)
9
1
−
(
−
10
)
=
1
−
(
−
10
)
9
1
=
(
10
)
9
+
1
{\displaystyle S_{9}=11\cdot {\frac {1-(-10)^{9}}{1-(-10)}}={\frac {1-(-10)^{9}}{1}}=(10)^{9}+1}
c
k
=
2
⋅
(
−
1
2
)
2
(
k
−
1
)
{\displaystyle c_{k}=2\cdot \left(-{\frac {1}{2}}\right)^{2(k-1)}}
Ze wzoru możemy w łatwy sposób wyliczyć kilka pierwszych wyrazów:
c
1
=
2
⋅
(
−
1
2
)
2
⋅
0
=
2
{\displaystyle c_{1}=2\cdot \left(-{\frac {1}{2}}\right)^{2\cdot 0}=2}
c
2
=
2
⋅
(
−
1
2
)
2
⋅
1
=
2
⋅
1
4
=
1
2
{\displaystyle c_{2}=2\cdot \left(-{\frac {1}{2}}\right)^{2\cdot 1}=2\cdot {\frac {1}{4}}={\frac {1}{2}}}
c
3
=
2
⋅
(
−
1
2
)
2
⋅
2
=
2
⋅
1
16
=
1
8
{\displaystyle c_{3}=2\cdot \left(-{\frac {1}{2}}\right)^{2\cdot 2}=2\cdot {\frac {1}{16}}={\frac {1}{8}}}
... Zatem widzimy, że
c
1
=
2
{\displaystyle c_{1}=2}
q
=
c
2
c
1
=
c
3
c
2
=
⋯
=
1
4
{\displaystyle q={\frac {c_{2}}{c_{1}}}={\frac {c_{3}}{c_{2}}}=\dots ={\frac {1}{4}}}
S
10
=
2
⋅
1
−
(
1
4
)
10
1
−
(
1
4
)
=
2
⋅
1
−
(
1
4
)
10
3
4
=
2
⋅
4
3
⋅
(
1
−
(
1
4
)
10
)
=
8
3
⋅
[
1
−
(
1
4
)
10
]
{\displaystyle S_{10}=2\cdot {\frac {1-\left({\frac {1}{4}}\right)^{10}}{1-\left({\frac {1}{4}}\right)}}=2\cdot {\frac {1-\left({\frac {1}{4}}\right)^{10}}{\frac {3}{4}}}=2\cdot {\frac {4}{3}}\cdot (1-\left({\frac {1}{4}}\right)^{10})={\frac {8}{3}}\cdot \left[1-\left({\frac {1}{4}}\right)^{10}\right]}
Wyznaczmy wzór ogólny na sumę n kolejnych elementów ciągu
(
d
n
)
{\displaystyle (d_{n})}
d
1
=
3
{\displaystyle d_{1}=3}
q
=
5
{\displaystyle q=5}
q
≠
1
{\displaystyle q\neq 1}
S
n
=
d
1
⋅
1
−
q
n
1
−
q
=
3
⋅
1
−
5
n
1
−
5
=
3
⋅
(
−
1
)
⋅
(
5
n
−
1
)
(
−
1
)
⋅
4
=
3
⋅
5
n
−
1
4
{\displaystyle S_{n}=d_{1}\cdot {\frac {1-q^{n}}{1-q}}=3\cdot {\frac {1-5^{n}}{1-5}}=3\cdot {\frac {(-1)\cdot (5^{n}-1)}{(-1)\cdot 4}}=3\cdot {\frac {5^{n}-1}{4}}}
Na koniec spróbujmy udowodnić, że ten wzór jest poprawny, nie odwołując się do wcześniej przedstawionego twierdzenia. Wypiszemy najpierw założenia i tezę, a potem przedstawimy dowód.
Założenia:
d
k
=
3
⋅
5
k
−
1
{\displaystyle d_{k}=3\cdot 5^{k-1}}
S
n
=
d
1
+
d
2
+
d
3
+
⋯
+
d
n
=
3
+
3
⋅
5
1
+
3
⋅
5
2
+
3
⋅
5
3
+
⋯
+
3
⋅
5
(
n
−
1
)
{\displaystyle S_{n}=d_{1}+d_{2}+d_{3}+\dots +d_{n}=3+3\cdot 5^{1}+3\cdot 5^{2}+3\cdot 5^{3}+\dots +3\cdot 5^{(n-1)}}
Teza:
S
n
=
3
⋅
5
n
−
1
4
{\displaystyle S_{n}=3\cdot {\frac {5^{n}-1}{4}}}
Dowód:
Sumę
S
n
=
3
+
3
⋅
5
1
+
3
⋅
5
2
+
3
⋅
5
3
+
⋯
+
3
⋅
5
(
n
−
1
)
{\displaystyle S_{n}=3+3\cdot 5^{1}+3\cdot 5^{2}+3\cdot 5^{3}+\dots +3\cdot 5^{(n-1)}}
q
=
5
{\displaystyle q=5}
5
S
n
=
3
⋅
5
1
+
3
⋅
5
2
+
3
⋅
5
3
+
⋯
+
3
⋅
5
n
{\displaystyle 5S_{n}=3\cdot 5^{1}+3\cdot 5^{2}+3\cdot 5^{3}+\dots +3\cdot 5^{n}}
Teraz odejmijmy od siebie obydwie sumy:
S
n
{\displaystyle S_{n}}
=
{\displaystyle =}
3
{\displaystyle 3}
+
{\displaystyle +}
3
⋅
5
1
{\displaystyle 3\cdot 5^{1}}
+
{\displaystyle +}
3
⋅
5
2
{\displaystyle 3\cdot 5^{2}}
+
{\displaystyle +}
3
⋅
5
3
{\displaystyle 3\cdot 5^{3}}
+
{\displaystyle +}
…
{\displaystyle \dots }
+
{\displaystyle +}
3
⋅
5
(
n
−
1
)
{\displaystyle 3\cdot 5^{(n-1)}}
−
{\displaystyle -}
5
S
n
{\displaystyle 5S_{n}}
=
{\displaystyle =}
3
⋅
5
1
{\displaystyle 3\cdot 5^{1}}
+
{\displaystyle +}
3
⋅
5
2
{\displaystyle 3\cdot 5^{2}}
+
{\displaystyle +}
3
⋅
5
3
{\displaystyle 3\cdot 5^{3}}
+
{\displaystyle +}
…
{\displaystyle \dots }
+
{\displaystyle +}
3
⋅
5
(
n
−
1
)
{\displaystyle 3\cdot 5^{(n-1)}}
+
{\displaystyle +}
3
⋅
5
n
{\displaystyle 3\cdot 5^{n}}
−
4
S
n
{\displaystyle -4S_{n}}
=
{\displaystyle =}
3
{\displaystyle 3}
+
{\displaystyle +}
0
{\displaystyle 0}
+
{\displaystyle +}
0
{\displaystyle 0}
+
{\displaystyle +}
0
{\displaystyle 0}
+
{\displaystyle +}
0
{\displaystyle 0}
+
{\displaystyle +}
0
{\displaystyle 0}
−
{\displaystyle -}
3
⋅
5
n
{\displaystyle 3\cdot 5^{n}}
−
4
S
n
=
3
−
3
⋅
5
n
=
3
(
1
−
5
n
)
{\displaystyle -4S_{n}=3-3\cdot 5^{n}=3(1-5^{n})}
-4 dochodzimy do:
S
n
=
3
⋅
1
−
5
n
−
4
=
3
⋅
5
n
−
1
4
{\displaystyle S_{n}=3\cdot {\frac {1-5^{n}}{-4}}=3\cdot {\frac {5^{n}-1}{4}}}
a co chcieliśmy udowodnić.
![{\displaystyle S_{n}={\frac {[2a_{1}+(n-1)\cdot r]\cdot n}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2864dc2a01d97ff08dfab8964de570f1f7d16fc)
![{\displaystyle S_{n}=a_{1}+a_{2}+\dots +a_{n}=[a_{1}+(1-1)\cdot r]+[a_{1}+(2-1)\cdot r]+\dots +[a_{1}+(n-1)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a4f62ca02067d36346f4aec03af31b2c0e4f903)
![{\displaystyle S_{n}=[a_{1}+(n-1)\cdot r]+[a_{1}+(n-2)\cdot r]+\dots +[a_{1}+(2-1)\cdot r]+[a_{1}+(1-1)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b18dc95191ea216dfe949b50b087d6873a0bff0)
![{\displaystyle [a_{1}+(1-1)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7e7dc0571522fcc0107ad7e45d5321f4606ac70)
![{\displaystyle [a_{1}+(2-1)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8488c03ea7a2b650dee3b6657e83bf998faa4091)
![{\displaystyle [a_{1}+(3-1)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4464416e8e892bb27bf904376d3b209eab1bb6ac)
![{\displaystyle [a_{1}+(n-1)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/326a371f14005d7b1fb59a74b860ed6d41be4d3a)
![{\displaystyle [a_{1}+(n-2)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d341ed847911c4e080d4d9c97d24f8a8be2a7cb)
![{\displaystyle [a_{1}+(n-3)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72691d7ff523e24156003f6a821c7800a0832933)
![{\displaystyle [a_{1}+(n-n)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c70407713c3870aa60b5739812525808585f4c3)
![{\displaystyle [2a_{1}+(n-1)\cdot r]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7364772054ed684a6ec4d972e221f6def188ab59)
![{\displaystyle 2S_{n}=[2a_{1}+(n-1)\cdot r]\cdot n}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f85bbbdd3b4b7cce0dda2d9ac7aced024c6f4d8d)
![{\displaystyle S_{10}=2\cdot {\frac {1-\left({\frac {1}{4}}\right)^{10}}{1-\left({\frac {1}{4}}\right)}}=2\cdot {\frac {1-\left({\frac {1}{4}}\right)^{10}}{\frac {3}{4}}}=2\cdot {\frac {4}{3}}\cdot (1-\left({\frac {1}{4}}\right)^{10})={\frac {8}{3}}\cdot \left[1-\left({\frac {1}{4}}\right)^{10}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dda98313130fbed08ddb2a947a2d4947cf01c5e4)
