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Ta có: ( Sửa đề )
\(A=4+4^2+4^3+...+4^{2021}+4^{2022}\)
\(A=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{2021}+4^{2022}\right)\)
\(A=20+4^2.\left(4+4^2\right)+...+4^{2020}.\left(4+4^2\right)\)
\(A=20+4^2.20+...+4^{2020}.20\)
\(A=20.\left(1+4^2+...+4^{2020}\right)\)
Vì \(20⋮20\) nên \(20.\left(1+4^2+...+4^{2020}\right)\)
Vậy \(A⋮20\)
\(#WendyDang\)
A = 2² + 2³ + 2⁴ + ... + 2²⁰²¹
⇒ 2A = 2³ + 2⁴ + 2⁵ + ... + 2²⁰²²
⇒ A = 2A - A
= (2³ + 2⁴ + 2⁵ + ... + 2²⁰²²) - (2² + 2³ + 2⁴ + ... + 2²⁰²¹)
= 2²⁰²² - 2²
= 2²⁰²² - 4
A= 22+23+24+25+...+22021
2A-A=23+24+25+...+22022
2A-A=(22+23+24+25+...+22021)-(23+24+25+...+22022)
A=22-22022
a)S = 1 + 2 + 22 + 23 + 24 +25 +26 +27 + 28 + 29
2S = 2.(1 + 2 + 22 + 23 + 24 +25 +26 +27 + 28 + 29)
2S = 2 + 22 + 23 + 24 +25 +26 +27 + 28 + 29 + 210
S = (2 + 22 + 23 + 24 +25 +26 +27 + 28 + 29 + 210) - (1 + 2 + 22 + 23 + 24 +25 +26 +27 + 28 + 29)
S = 210 - 1
Suy ra: S = \(\frac{2^{9+1}-1}{2-1}\)
S = \(\frac{2^{10}-1}{1}\)
S = 210 - 1
S = 1023
b)Mình không thể giúp bạn vì mình không rõ 5.28 hay (5.2)8
a) \(A=2+2^2+...+2^{2024}\)
\(2A=2^2+2^3+...+2^{2025}\)
\(2A-A=2^2+2^3+...+2^{2025}-2-2^2-...-2^{2024}\)
\(A=2^{2025}-2\)
b) \(2A+4=2n\)
\(\Rightarrow2\cdot\left(2^{2025}-2\right)+4=2n\)
\(\Rightarrow2^{2026}-4+4=2n\)
\(\Rightarrow2n=2^{2026}\)
\(\Rightarrow n=2^{2026}:2\)
\(\Rightarrow n=2^{2025}\)
c) \(A=2+2^2+2^3+...+2^{2024}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2023}+2^{2024}\right)\)
\(A=2\cdot3+2^3\cdot3+...+2^{2023}\cdot3\)
\(A=3\cdot\left(2+2^3+...+2^{2023}\right)\)
d) \(A=2+2^2+2^3+...+2^{2024}\)
\(A=2+\left(2^2+2^3+2^4\right)+\left(2^5+2^6+2^7\right)+...+\left(2^{2022}+2^{2023}+2^{2024}\right)\)
\(A=2+2^2\cdot7+2^5\cdot7+...+2^{2022}\cdot7\)
\(A=2+7\cdot\left(2^2+2^5+...+2^{2022}\right)\)
Mà: \(7\cdot\left(2^2+2^5+...+2^{2022}\right)\) ⋮ 7
⇒ A : 7 dư 2
Sai đề câu E sửa lại 95 hoặc 93 vì đây là dãy số mũ lẻ. Ta có :
\(E=3+3^3+3^5+3^7+...+3^{95}\)
\(\Rightarrow\) \(9E=3^3+3^5+3^7+3^9+...+3^{95}+3^{97}\)
\(\Rightarrow\) \(8E=3^{97}-3\)
\(\Rightarrow\) \(E=\frac{3^{97}-3}{8}\)
\(E=3+3^3+3^5+3^7+.......+3^{95}\)
\(\Rightarrow9E=3^3+3^5+3^7+3^9+...+3^{97}\)
\(\Rightarrow9E-E=\left(3^3+3^5+3^7+3^9+....+3^{97}\right)-\left(3+3^3+3^5+3^7+.....+3^{95}\right)\)
\(\Rightarrow8E=3^{97}-3\)
\(\Rightarrow E=\frac{3^{97}-3}{8}\)
\(F=1+2018+2018^2+......+2018^{2017}\)
\(=2018^0+2018^1+2018^2+....+2018^{2017}\)
\(\Rightarrow2018F=2018^1+2018^2+2018^3+....+2018^{2018}\)
\(\Rightarrow2018F-F=\left(2018^1+2018^2+2018^3+....+2018^{2018}\right)-\left(2018^0+2018^1+2018^2+....+2018^{2017}\right)\)
\(\Rightarrow2017F=2018^{2018}-1\)
\(\Rightarrow F=\frac{2018^{2018}-1}{2017}\)
Ta có :
B = 2100 - 299 + 298 - 297 + ... + 22 - 2 + 1
=> B = ( 2100 + 298 + ... + 22 + 1 ) - ( 299 + 297 + ... + 2 )
=> 22B = 2 . [ ( 2100 + 298 + ... + 22 + 1 ) - ( 299 + 297 + ... + 2 ) ]
=> 4B = ( 2102 + 2100 + ... + 22 ) - ( 2101 + 299 + ... + 23 )
=> 4B - B = [( 2102 + 2100 + ... + 22 ) - ( 2101 + 299 + ... + 23 )] - [( 2100 + 298 + ... + 22 + 1 ) - ( 299 + 297 + ... + 2 )]
=> 3B = ( 2102 - 1 ) + ( 2 - 2101 )
=> 3B = 2101 - 1
=> B = \(\frac{2^{101} - 1}{3}\)
gọi dãy số là A, ta có:
A = 2100 - 299 - ...... - 21
2A = 2101 - 2100 - .... - 22
2A = ( 2101 - ... - 22 ) - ( 2100 - ... - 2 )
A = 2101 - 2
A=2^2018 nha bạn