Cho D = 7+73+75+...+71999
a, Tính D
b, CMR D chia hết cho 35
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A=(7+73)+(75+77)+....+(71997+71999)
A=7.(1+72)+75.(1+72)+....+71997.(1+72)
A=7.50+75.50+79.50+.....+71997.50
=>A chia hết cho 5 (1)
A=(7+73+75+....+71999)=7.(70+72+74+....+71998)
=>A chia hết cho 7 (2)
Mà ƯCLN(5;7)=1=>A chia hết cho 35
a) D = 7+73+75+...+71999
=> 72D= 73+75+...+71999+72001
=> 72D-D=(73+75+...+71999+72001)-( 7+73+75+...+71999)
=> 72D-D hay D(72-1)=48D=72001-7
=> D=(72001-7)/48
a, D = 7+73+75+.....+71999
72D = 73+75+77+.....+72001
48D = 72D - D = 72001-7
=> D = \(\frac{7^{2001}-7}{48}\)
b, D = 7+73+75+.....+71999
D = (7+73)+(75+77)+.....+(71997+71999)
D = 1(7+73)+74(7+73)+.....+71996(7+73)
D = 1.350 + 74.350+.....+71996.350
D = 350(1+74+......+71996) chia hết cho 350
=> D chia hết cho 35 ( Vì 350 chia hết cho 35)
a: \(=2^2\left(1+2\right)+2^4\left(1+2\right)=3\left(2^2+2^4\right)⋮3\)
b: \(=4^{20}\left(1+4\right)+4^{22}\left(1+4\right)=5\left(4^{20}+4^{22}\right)⋮5\)
c: \(A=\left(1+4+4^2\right)+...+4^{96}\left(1+4+4^2\right)\)
\(=21\left(1+...+4^{96}\right)⋮21\)
d: \(B=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{35}\left(1+7\right)\)
\(=8\left(7+7^3+...+7^{35}\right)⋮8\)
\(B=7\left(1+7+7^2\right)+...+7^{34}\left(1+7+7^2\right)\)
\(=57\left(7+...+7^{34}\right)\) chia hếtcho 3 và 19
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