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Nếu B chia hết cho 21 suy ra B chia hết cho 3,7
B=(2+2^2)+(2^3+2^4)+...+(2^29+2^30)
=2(1+2)+2^3(1+2)+...+2^29(1+2) =2.3+2^3.3+...+2^29.3
=3(2+2^3+...+2^29) chia hết cho 3
B=(2+2^2+2^3)+...+(2^28+2^29+2^30)
=2(1+2+2^2)+...+2^28(1+2+2^2)
=2.7+...+2^28.7 =7(2+...+2^28) chia hết cho 7 Vậy B chia hết cho 21
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(M=21^9+21^8+21^7+...+21+1\)
Do \(21^n\) luôn có tận cùng là 1
\(\Rightarrow M=21^9+21^8+21^7+...+21+1\)
Tân cùng của M là:
\(1+1+1+1+1+1+1+1+1+1=10\) tận cùng là 0
\(\Rightarrow M⋮10\)
\(\Leftrightarrow M⋮2;5\)
b) \(N=6+6^2+6^3+...+6^{2020}\)
\(N=6.\left(1+6\right)+6^3.\left(1+6\right)+...+6^{2019}.\left(1+6\right)\)
\(N=6.7+6^3.7+...+6^{2019}.7\)
\(N=7.\left(6+6^3+...+6^{2019}\right)⋮7\)
\(\Rightarrow N⋮7\)
Ta thấy: \(N=6+6^2+6^3+...+6^{2020}⋮6\)
Mà \(6⋮̸9\)
\(\Rightarrow N⋮̸9\)
c) \(P=4+4^2+4^3+...+4^{23}+4^{24}\)
\(P=1.\left(4+4^2\right)+4^2.\left(4+4^2\right)+...+4^{20}.\left(4+4^2\right)+4^{22}.\left(4+4^2\right)\)
\(P=1.20+4^2.20+...+4^{20}.20+4^{22}.20\)
\(P=20.\left(1+4^2+...+4^{20}+4^{22}\right)⋮20\)
\(\Rightarrow P⋮20\)
\(P=4+4^2+4^3+...+4^{23}+4^{24}\)
\(P=4.\left(1+4+4^2\right)+...+4^{22}.\left(1+4+4^2\right)\)
\(P=4.21+...+4^{22}.21\)
\(P=21.\left(4+...+4^{22}\right)⋮21\)
\(\Rightarrow P⋮21\)
d) \(Q=6+6^2+6^3+...+6^{99}\)
\(Q=6.\left(1+6+6^2\right)+...+6^{97}.\left(1+6+6^2\right)\)
\(Q=6.43+...+6^{97}.43\)
\(Q=43.\left(6+...+6^{97}\right)⋮43\)
\(\Rightarrow Q⋮43\)
Chúc bạn học tốt!
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Bài 1:
\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)
Bài 2:
\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)
Bài 1 :
\(2^{49}=\left(2^7\right)^7=128^7\)
\(5^{21}=\left(5^3\right)^7=125^7\)
mà \(125^7< 128^7\)
\(\Rightarrow2^{49}>5^{21}\)
Bài 2 :
a) \(S=1+3+3^2+3^3+...3^{99}\)
\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)
\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)
\(\Rightarrow dpcm\)
b) \(S=1+4+4^2+4^3+...4^{62}\)
\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)
\(\Rightarrow S=21+4^3.21+...4^{60}.21\)
\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)
\(\Rightarrow dpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
A=1+4+42+...+459A=1+4+42+...+459
A=(1+4+42)+(43+44+45)+...+(457+458+459)A=(1+4+42)+(43+44+45)+...+(457+458+459)
A=(1+4+42)+43(1+4+42)+...+447(1+4+42)A=(1+4+42)+43(1+4+42)+...+447(1+4+42)
A=21+43.21+...+447.21A=21+43.21+...+447.21
A=21(1+43+...+447)A=21(1+43+...+447)
⇒A⋮21
các số như 43,447,459,458........ là 4 mũ và các số đằng sau là số mũ
câu b cũng làm như vậy nhưng dổi các số và kết quả
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=2+2^2+2^3+...+2^60 = 2(1+2+2^2+2^3) + 2^5(1+2+2^2+2^3) + ... + 2^57(1+2+2^2+2^3)
A=(2+2^5+...+2^57)*15 chia het cho 15
CM:
A chia hết cho 21
=> A chia hết cho 3 và 7
Ta có
A=2(1+2)+2^3(1+2)+..............+2^59(1...
A=3(2+2^3+2^5+........+2^59)chia hết cho 3
Ta có :
A=2(1+2+2^2)+2^4(1+2+2^2)+...........+2...
A=7(2+2^4+2^7+..........+2^58)
=> A chia hết cho 3 và 7=> A chia hết
Vậy A chia hết cho 21 và 15
Nếu B chia hết cho 21 suy ra B chia hết cho 3,7
B=(2+2^2)+(2^3+2^4)+...+(2^29+2^30)
=2(1+2)+2^3(1+2)+...+2^29(1+2)
=2.3+2^3.3+...+2^29.3
=3(2+2^3+...+2^29) chia hết cho 3
B=(2+2^2+2^3)+...+(2^28+2^29+2^30)
=2(1+2+2^2)+...+2^28(1+2+2^2)
=2.7+...+2^28.7
=7(2+...+2^28) chia hết cho 7
Vậy B chia hết cho 21
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`#3107.101107`
a,
\(C=2+2^3+2^5+...+2^{23}\)
\(=\left(2+2^3+2^5\right)+\left(2^5+2^7+2^9\right)+...+\left(2^{19}+2^{21}+2^{23}\right)\)
\(=2\left(1+2^2+2^4\right)+2^5\cdot\left(1+2^2+2^4\right)+...+2^{19}\cdot\left(1+2^2+2^4\right)\)
\(=\left(1+2^2+2^4\right)\cdot\left(2+2^5+...+2^{19}\right)\)
\(=21\cdot\left(2+2^5+...+2^{19}\right)\)
Vì \(21\text{ }⋮\text{ }21\)
\(\Rightarrow21\left(2+2^5+...+2^{19}\right)\text{ }⋮\text{ }21\)
Vậy, \(C\text{ }⋮\text{ }21\)
b,
\(C=2+2^3+2^5+...+2^{23}\)
\(=\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{21}+2^{23}\right)\)
\(=\left(2+2^3\right)+2^4\cdot\left(2+2^3\right)+...+2^{20}\cdot\left(2+2^3\right)\)
\(=\left(2+2^3\right)\cdot\left(1+2^4+...+2^{20}\right)\)
\(=10\cdot\left(1+2^4+...+2^{20}\right)\)
Vì \(10\text{ }⋮\text{ }10\)
\(\Rightarrow10\cdot\left(1+2^4+...+2^{20}\right)\text{ }⋮\text{ }10\)
Vậy, \(C\text{ }⋮\text{ }10.\)
a) c = 2 + 2³ + 2⁵ + ... + 2¹⁹ + 2²¹ + 2²³
= (2 + 2³ + 2⁵) + (2⁷ + 2⁹ + 2¹¹) + ... + (2¹⁹ + 2²¹ + 2²³)
= 2.(1 + 2² + 2⁴) + 2⁷.(1 + 2² + 2⁴) + ... + 2¹⁹.(1 + 2² + 2⁴)
= 2.21 + 2⁷.21 + ... + 2¹⁹.21
= 21.(2 + 2⁷ + ... + 2¹⁹) ⋮ 21
Vậy c ⋮ 21
b) c = 2 + 2³ + 2⁵ + 2⁷ + ... + 2²¹ + 2²³
= (2 + 2³) + (2⁵ + 2⁷) + ... + (2²¹ + 2²³)
= 10 + 2⁴.(2 + 2³) + ... + 2²⁰.(2 + 2³)
= 10 + 2⁴.10 + ... + 2²⁰.10
= 10.(1 + 2⁴ + ... + 2²⁰) ⋮ 10
Vậy c ⋮ 10
\(\left(2+2^3+2^5\right)+\left(2^2+2^4+2^6\right)+.........\)
\(2\left(1+2^2+2^4\right)+2^2\left(1+2^2+2^4\right)\)+...
\(2\left(21\right)+2^2\left(21\right)+....\)
21(2+2^2+...)
vậy
Nếu B chia hết cho 21 suy ra B chia hết cho 3,7
B=(2+2^2)+(2^3+2^4)+...+(2^29+2^30)
=2(1+2)+2^3(1+2)+...+2^29(1+2)
=2.3+2^3.3+...+2^29.3 =3(2+2^3+...+2^29) chia hết cho 3
B=(2+2^2+2^3)+...+(2^28+2^29+2^30)
=2(1+2+2^2)+...+2^28(1+2+2^2) =2.7+...+2^28.7
=7(2+...+2^28) chia hết cho 7 Vậy B chia hết cho 21