biblioteka.by/andrei_62
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Вacилий С.
Мінск, Belarus
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Headline:
Magnetic self-motion
First Name:
Andrei
Last Name:
Verner
Country:
Kyrgyzstan
Sex:
Man
City:
Frunze
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All posts · Posts by Andrei Verner · Posts by others · RSS · Subscribe
 
Time range (days ago):
31.12.2025
Andrei Verner edited his profile info
166 days ago
30.12.2025
Andrei Verner edited his profile info
167 days ago
19.06.2025
Andrei Verner edited his profile info
361 days ago
Andrei Verner edited his profile info
361 days ago
18.05.2025
Andrei Verner edited his profile info
393 days ago
02.03.2025
Andrei Verner edited his profile info
470 days ago
07.10.2024
Andrei Verner edited his profile info
616 days ago
05.10.2024
Andrei Verner added new article
Суммы прогрессий: 1,2,3,4,5..., -1,-2,-3,-4,-5... Можно найти с помощью формулы:Sn= (n²a₁+n)/2. Суммы прогрессий: 1,3,6,10,15..., -1,-3,-6,-10,-15... Можно найти с помощью формулы:Sn= ((n+a₁)³-(n+a₁))/6. Суммы прогрессий: 1,4,9,16,25..., -1,-4,-9,-16,-25... Можно найти с помощью формулы:Sn= a₁(n+a₁)(n²a₁+0.5n)/3. (где n - количество суммируемых членов, a₁ -первый член прогрессии).
618 days ago
Andrei Verner added new article
Progress Sums: 1,2,3,4,5..., -1,-2,-3,-4,-5... It can be found using the formula: Sn=(n²a₁+n)/2. Progress Sum: 1,3,6,10,15..., -1,-3,-6,-10,-15... It can be found using the formula: Sn= ((n+a₁)³-(n+a₁))/6. Progress Sum: 1,4,9,16,25..., -1,-4,-9,-16,-25... It can be found using the formula: Sn= a₁(n+a₁)(n²a₁+0.5n)/3. (Where n - is the number of summable terms, a₁ - is the first term of the progression).
618 days ago
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Суммы прогрессий: 1,2,3,4,5..., -1,-2,-3,-4,-5... Можно найти с помощью формулы:Sn= (n²a₁+n)/2. Суммы прогрессий: 1,3,6,10,15..., -1,-3,-6,-10,-15... Можно найти с помощью формулы:Sn= ((n+a₁)³-(n+a₁))/6. Суммы прогрессий: 1,4,9,16,25..., -1,-4,-9,-16,-25... Можно найти с помощью формулы:Sn= a₁(n+a₁)(n²a₁+0.5n)/3. (где n - количество суммируемых членов, a₁ -первый член прогрессии).
618 days ago · From Andrei Verner
Progress Sums: 1,2,3,4,5..., -1,-2,-3,-4,-5... It can be found using the formula: Sn=(n²a₁+n)/2. Progress Sum: 1,3,6,10,15..., -1,-3,-6,-10,-15... It can be found using the formula: Sn= ((n+a₁)³-(n+a₁))/6. Progress Sum: 1,4,9,16,25..., -1,-4,-9,-16,-25... It can be found using the formula: Sn= a₁(n+a₁)(n²a₁+0.5n)/3. (Where n - is the number of summable terms, a₁ - is the first term of the progression).
618 days ago · From Andrei Verner
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