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\(A=5+5^2+5^3+....+5^9+5^{10}\)
\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^9+5^{10}\right)\)
\(A=5\left(1+5\right)+5^3\left(1+5\right)+...+5^9\left(1+5\right)\)
\(A=5.6+5^3.6+....+5^9.6\)
\(A=6\left(5+5^3+....+5^9\right)\)
vì \(6⋮6\Rightarrow A=6\left(5+5^3+....+5^9\right)\)
\(\Rightarrow A⋮6\)
chúc bạn học giỏi ^^
Ta có :
A=5+52+53...+59+510
A = ( 5 + 52 ) + ( 53 + 54 ) + ... + ( 59 + 510 )
A = 5 . (1 + 5 ) + 53 . ( 1 + 5 ) + ... + 59 . ( 1 + 5 )
A = 5 . 6 + 53 . 6 + ... + 59 . 6
A = 6 . ( 5 + 53 + ... + 59 ) \(⋮\)6
Vậy ...
\(3,1+5^2+5^4+...+5^{26}\)
\(=\left(1+5^2\right)+\left(5^4+5^6\right)+...+\left(5^{24}+5^{26}\right)\)
\(=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{24}\left(1+5^2\right)\)
\(=26+5^4.26+...+5^{24}.26\)
\(=26\left(5^4+...+5^{24}\right)\)
Vì \(26⋮26\)
\(\Rightarrow26\left(5^4+...+5^{24}\right)⋮26\)
\(\Rightarrow1+5^2+5^4+...+5^{26}⋮26\)
\(4,1+2^2+2^4+...+2^{100}\)
\(=\left(1+2^2+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=\left(1+2^2+2^4\right)+....+2^{98}\left(1+2^2+2^4\right)\)
\(=21+2^6.21...+2^{98}.21\)
\(=21\left(2^6+...+2^{98}\right)\)
Có : \(21\left(2^6+...+2^{98}\right)⋮21\)
\(\Rightarrow1+2^2+2^4+...+2^{100}⋮21\)
Ta có :
\(A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\) \(=30+30.2^5+...30.2^{17}\)
\(=30.\left(1+2^5+...+2^{17}\right)\)
\(=5.6.\left(1+2^5+...+2^{17}\right)\) chia hết cho 5
Vậy A chia hết cho 5
\(A=2+2^2+2^3+...+2^{20}\)
\(A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(A=\left(2.1+2.2+2.2^2+2.2^3\right)+\left(2^5.1+2^5.2+2^5.2^2+2^5.2^3\right)+...\left(2^{17}.1+2^{17}.2+2^{17}.2^2+2^{17}.2^3\right)\)
\(A=2.\left(1+2+4+8\right)+2^5.\left(1+2+4+8\right)+...+2^{17}.\left(1+2+4+8\right)\)
\(A=2.15+2^5.15+...+2^{17}.15\)
\(A=15.\left(2+2^5+...+2^{17}\right)\)
Vì 15 chia hết cho 5
=> A chia hết cho 5
A=2.(1+2+4+8)+...2^17(1+2+4+8)
A=2.15+2^5.15+...+2^17.15
A=15.(2+2^5+...+2^17) chia het cho 5
Vay.............