Thu gọn C, biết :
C = 32023 - 32022 + 32021 - 32020 + 32019 - ... - 32 + 3.
Giúp mình với!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=3^{2024}-3^{2023}+3^{2022}-3^{2021}+...+3^2-3\)
\(3S=3^{2025}-3^{2024}+3^{2023}-3^{2022}+...+3^3-3^2\)
\(3S+S=3^{2025}-3^{2024}+3^{2023}-3^{2022}+...+3^3-3^2+3^{2024}-3^{2023}+3^{2022}-3^{2021}+...+3^2-3\)\(4S=3^{2025}-3\)
\(S=\dfrac{3^{2025}-3}{4}\)
S = 32024 - 32023 + 32022 - 32021 +... + 32 - 3
3.S = 32025 - 32024 + 32022 -32021 + ....+ 33 - 32
3S + S = 32025 - 32024 + 32022 - 32021 +...+33 - 32+(32024-32023+...-3)
4S = 32025 - 32024 + 32022 - 32021+...+33-32 + 32024-32023+...-3
4S = 32025 - (32024 - 32024) -...-(32 - 32) - 3
4S = 32025 - 3
S = \(\dfrac{3^{2025}-3}{4}\)
Ta có: \(B=1+3+3^2+\cdots+3^{2021}\)
=>\(3B=3+3^2+3^3+\cdots+3^{2022}\)
=>3B-B=\(3+3^2+3^3+\cdots+3^{2022}-1-3-\cdots-3^{2021}\)
=>2B=\(3^{2022}-1\)
=>\(B=\frac{3^{2022}-1}{2}\)
C-B\(=\frac{3^{2022}}{2}-\frac{3^{2022}-1}{2}=\frac12\)
\(\Rightarrow12x-33=3\\ \Rightarrow12x=36\\ \Rightarrow x=3\)
A = 1 + 3 + 3² + ... + 3²⁰²³
⇒ 3A = 3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴
⇒ 2A = 3A - A
= (3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴) - (1 + 3 + 3² + ... + 3²⁰²³)
= 3²⁰²⁴ - 1
⇒ A = (3²⁰²⁴ - 1) : 2
⇒ A < B
A=1+3+32+33+34+........+32022+32023
3A=3+32+33+............+32023+32024
3A-A=(3+32+33+..........+32023+32024
A =1+3+32+.....+32022+32023
3.A =3+32+33+.....+32023+32024
3.A -A=(3+32+33+.....+32023+32024 ) - (1+3+32+.....+32022+32023)
2A =32024-1
A =\(\dfrac{3^{2024}-1}{2}\)
A = 1 + 3 + 32 + 33 + 34 + ... + 32022
3A = 3 + 32 + 33 + ... + 34 + ... + 32022 + 32023
3A - A = (3 + 32 + 33 + ... + 34 + 32022 + 32023) - (1 + 3+...+ 32022)
2A = 3 + 32 + 33 + 34 + ... + 32022 + 32023 - 1 - 3 - ... - 32022
2A = (3 - 3) + (32 - 32) + (34 - 34) + (32022 - 32022) + (32023 - 1)
2A = 32023 - 1
A = \(\dfrac{3^{2023}-1}{2}\)
A = \(\dfrac{3^{2023}}{2}\) - \(\dfrac{1}{2}\)
B - A = \(\dfrac{3^{2023}}{2}\) - (\(\dfrac{3^{2023}}{2}\) - \(\dfrac{1}{2}\))
B - A = \(\dfrac{3^{2023}}{2}\) - \(\dfrac{3^{2023}}{2}\) + \(\dfrac{1}{2}\)
B - A = \(\dfrac{1}{2}\)
Lời giải:
$A=1+3+3^2+3^3+...+3^{2021}$
$3A=3+3^2+3^3+...+3^{2022}$
$\Rightarrow 3A-A=(3+3^2+3^3+...+3^{2022}) - (1+3+3^2+3^3+...+3^{2021})$
$\Rightarrow 2A=3^{2022}-1$
$\Rightarrow A=\frac{3^{2022}-1}{2}$
$B-A=\frac{3^{2022}}{2}-\frac{3^{2022}-1}{2}=\frac{1}{2}$
`#3107`
\(\left(3^{2021}+3^{2020}\right)\div3^{2020}\\ =3^{2021}\div3^{2020}+3^{2020}\div3^{2020}\\ =3^{2021-2020}+3^{2020-2020}\\ =3+1=4\)
Ta có: \(5x-\left(3^{2021}-3^{2019}\right):3^{2017}=27\)
=>\(5x-3^4+3^2=27\)
=>5x-81+9=27
=>5x=27-9+81=18+81=99
=>x=99/5(loại)
Trường nào đó?