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\(S=3^{2024}-3^{2023}+3^{2022}-3^{2021}+...+3^2-3\)
\(3S=3^{2025}-3^{2024}+3^{2023}-3^{2022}+...+3^3-3^2\)
\(3S+S=3^{2025}-3^{2024}+3^{2023}-3^{2022}+...+3^3-3^2+3^{2024}-3^{2023}+3^{2022}-3^{2021}+...+3^2-3\)\(4S=3^{2025}-3\)
\(S=\dfrac{3^{2025}-3}{4}\)
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 = \(\dfrac{3^{2025}-3}{4}\)
\(\Rightarrow12x-33=3\\ \Rightarrow12x=36\\ \Rightarrow x=3\)
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:
$A=1+3+3^2+3^3+...+3^{2021}$
$3A=3+3^2+3^3+...+3^{2022}$
$\Rightarrow 3A-A=(3+3^2+3^3+...+3^{2022}) - (1+3+3^2+3^3+...+3^{2021})$
$\Rightarrow 2A=3^{2022}-1$
$\Rightarrow A=\frac{3^{2022}-1}{2}$
$B-A=\frac{3^{2022}}{2}-\frac{3^{2022}-1}{2}=\frac{1}{2}$
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