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LK
3
11 tháng 9 2019
B chia hết cho 30 :
B = 5 + 52 + ... + 596
B = ( 5 + 52 ) + ( 53 + 54 ) + ... + ( 595 + 596 )
B = 5 ( 1 + 5 ) + 53 ( 1 + 5 ) + ... + 595 ( 1 + 5 )
B = 5 . 6 + 53 . 6 + ... + 595 . 6
B = 6 ( 5 + 53 + ... + 595 )
= > B chia hết cho 6
Vì B các số hạng của B là những số chia hết cho 5 ( 5 ; 52 ; ... ; 596 )
= > Tổng B chia hết cho 5
Vì ( 5 ; 6 ) = 1 = > B chia hết cho 30
31 tháng 10 2023
Ta có \(B=5^{2024}+5^{2023}+5^{2022}\)
\(B=5^{2022}\left(5^2+5+1\right)\)
\(B=31.5^{2022}⋮31\)
Vậy \(B⋮31\) (đpcm)
27 tháng 12 2017
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28 tháng 12 2017
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2 tháng 10 2016
Ta có A = \(1+5+5^2+...+5^{2015}\)
=> 5A = \(5+5^2+5^3+...+5^{2016}\)
=> 5A - A = \(5+5^2+5^3+...+5^{2016}-1-5-5^2-...-5^{2015}\)
=> 4A = \(5^{2016}-1\)
=> A = \(\left(5^{2016}-1\right):4\)
=> A chia hết cho 31
NL
3
9 tháng 7 2021
\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=\)
\(=5^3.21⋮7\)
22 tháng 9 2017
a/ Ta có :
\(9^{1945}-2^{1930}=\left(9^5\right)^{389}-\left(2^{10}\right)^{193}=\left(.....9\right)-\left(.....4\right)=\left(............5\right)⋮5\)
\(\Leftrightarrowđpcm\)
\(B=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{2023}+5^{2024}\right)\\ =5\left(1+5\right)+5^3\left(1+5\right)+...+5^{2023}\left(1+5\right)\\ =5.6+5^3.6+...+5^{2023}.6\\ =6.\left(5+5^3+...+5^{2023}\right)⋮6\)
\(B=5+5^2+...+5^{2024}=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{2023}\left(1+5\right)\)
\(=6\left(5+5^3+...+5^{2023}\right)⋮6\)
Ta có đpcm
\(B=5\left(1+5+5^2\right)+...+5^{2022}\left(1+5+5^2\right)=21\left(5+...+5^{2022}\right)⋮21\)
Vậy ta có đpcm
\(B=5\left(1+5+5^2+5^3\right)+...+5^{2021}\left(1+5+5^2+5^3\right)\)
\(=126\left(5+...+5^{2021}\right)⋮126\)
Vậy ta có đpcm
sửa đề chia hết cho 126 nhé