Cho A = 5 + 52 + 53 + … + 52022. Tìm x để 4A + 5 = 5x
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a) \(B=5+5^2+5^3+...+5^{2022}\)
\(\Rightarrow5B=5^2+5^3+5^4+...+5^{2023}\)
\(\Rightarrow4B=5^{2023}-5\)
b) \(4B+5=5^X\)
Hay \(5^{2023}-5+5=5^X\)
\(5^{2023}=5^x\)
\(\Rightarrow x=2023\)
B = 5 + 52 + 53 +...+ 52022
5.B = 52 + 53 +....+ 52023
5B- B = 52023 - 5
4B = 52023 - 5
b, 4B + 5 = 5\(^x\) ⇒ 52023 - 5 + 5 = 5\(^x\)
5\(^{2023}\) = 5\(x\)
\(x\) = 2023
a) Ta có A = 21 + 22 + 23 + ... + 22022
2A = 22 + 23 + 24 + ... + 22023
2A - A = ( 22 + 23 + 24 + ... + 22023 ) - ( 21 + 22 + 23 + ... + 22022 )
A = 22023 - 2
Lại có B = 5 + 52 + 53 + ... + 52022
5B = 52 + 53 + 54 + ... + 52023
5B - B = ( 52 + 53 + 54 + ... + 52023 ) - ( 5 + 52 + 53 + ... + 52022 )
4B = 52023 - 5
B = \(\dfrac{5^{2023}-5}{4}\)
b) Ta có : A + 2 = 2x
⇒ 22023 - 2 + 2 = 2x
⇒ 22023 = 2x
Vậy x = 2023
Lại có : 4B + 5 = 5x
⇒ 4 . \(\dfrac{5^{2023}-5}{4}\) + 5 = 5x
⇒ 52023 - 5 + 5 = 5x
⇒ 52023 = 5x
Vậy x = 2023
5A=5+5^2+...+5^2023
=>4A=5^2023-1
=>\(A=\dfrac{5^{2023}-1}{4}\)
\(2B-A=\dfrac{5^{2023}}{4}-\dfrac{5^{2023}-1}{4}=\dfrac{1}{4}\)
\(S=5+5^2+5^3+...+5^{2020}+5^{2021}\)
=>\(5\cdot S=5^2+5^3+5^4+...+5^{2021}+5^{2022}\)
=>\(5S-S=5^2+5^3+...+5^{2021}+5^{2022}-5-5^2-5^3-...-5^{2020}-5^{2021}\)
=>\(4S=5^{2022}-5\)
=>\(4S+5=5^{2022}\)
a) \(S=1+2+2^2+..+2^{2022}\)
\(2S=2+2^2+2^3+...+2^{2023}\)
\(2S-S=2+2^2+2^3+...+2^{2023}-1-2-2^2-...-2^{2022}\)
\(S=2^{2023}-1\)
b) \(S=3+3^2+3^3+...+3^{2022}\)
\(3S=3^2+3^3+...+3^{2023}\)
\(3S-S=3^2+3^3+....+3^{2023}-3-3^2-...-3^{2022}\)
\(2S=3^{2023}-3\)
\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)
c) \(S=4+4^2+4^3+...+4^{2022}\)
\(4S=4^2+4^3+...+4^{2023}\)
\(4S-S=4^2+4^3+...+4^{2023}-4-4^2-...-4^{2022}\)
\(3S=4^{2023}-4\)
\(S=\dfrac{4^{2023}-4}{3}\)
d) \(S=5+5^2+...+5^{2022}\)
\(5S=5^2+5^3+...+5^{2023}\)
\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)
\(4S=5^{2023}-5\)
\(S=\dfrac{5^{2023}-5}{4}\)
A= 1 + 5 + 52 + 5 3 + ... + 5800
5A= 5 + 52 + 53 + .... +5 800 + 5801
5A - A = 5801 - 1
4a = 5801 - 1
5801 - 1 +1 = 5n
⇒ 5801 = 5n ⇒ n = 801