Home
О нас
Products
Services
Регистрация
Войти
Поиск
ТакоГоНикаТочноНеТ
@ТакоГоНикаТочноНеТ
July 2022
1
2
Report
Дана геометрическая прогрессия (bn). Найдите b1,q,S5, если bn= 3
-----------
3(в степени 2-n)
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms of service
You must agree before submitting.
Send
Answers & Comments
freepvps
Verified answer
Bn = 3 / 3^(2-n) = 3 * 3^(n-2) = 3^(n-1)
=> b1 = 3 ^ 0 = 1
=> b2 = b1 * 3
=> q = 3
=> s5 = b1 * (1-q^5)/1-q = (1-243)/-2 = 121
16 votes
Thanks 61
рекомендуемые вопросы
rarrrrrrrr
August 2022 | 0 Ответы
o chem dolzhny pozabotitsya v pervuyu ochered vzroslye pri organizacionnom vyvoze n
danilarsentev
August 2022 | 0 Ответы
est dva stanka na kotoryh vypuskayut odinakovye zapchasti odin proizvodit a zapcha
myachina8
August 2022 | 0 Ответы
najti po grafiku otnoshenie v3v1 v otvetah napisano 9 no nuzhno reshenie
ydpmn7cn6w
August 2022 | 0 Ответы
Choose the correct preposition: 1.I am fond (out,of,from) literature. 2.where ar...
millermilena658
August 2022 | 0 Ответы
opredelite kak sozdavalas i kto sozdaval arabskoe gosudarstvo v kracii
MrZooM222
August 2022 | 0 Ответы
ch ajtmanov v rasskaze krasnoe yabloko ispolzuet metod rasskaz v rasskaze opi
timobila47
August 2022 | 0 Ответы
kakovo bylo naznachenie kazhdoj iz chastej vizantijskogo hrama pomogite pozhalujsta
ivanyyaremkiv
August 2022 | 0 Ответы
moment. 6....
pozhalujsta8b98a56c0152a07b8f4cbcd89aa2f01e 97513
sarvinozwakirjanova
August 2022 | 0 Ответы
pomogite pozhalusto pzha519d7eb8246a08ab0df06cc59e9dedb 6631
×
Report "Дана геометрическая прогрессия (bn). Найдите b1,q,S5, если bn= 3 ..."
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
О нас
Политика конфиденциальности
Правила и условия
Copyright
Контакты
Helpful Social
Get monthly updates
Submit
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
Bn = 3 / 3^(2-n) = 3 * 3^(n-2) = 3^(n-1)=> b1 = 3 ^ 0 = 1
=> b2 = b1 * 3
=> q = 3
=> s5 = b1 * (1-q^5)/1-q = (1-243)/-2 = 121