a=2^1 +2^2 +...+2^60
chứng minh rằng achia hết cho 14
1, Chứng minh rằng 2139+3921 chia hết cho 45
2, Cho A =2+22+23+.....+260
Chứng minh rằng Achia hết cho 3;7;15
ChoA=2+2^2+2^3+...+2^60. Chứng minh rằng Achia hết cho 3,5và 7
\(A=2+2^2+...+2^{60}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)
\(A=3\cdot\left(2+2^3+...+2^{59}\right)\)
Vậy A chia hết cho 3
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\(A=2+2^2+2^3+...+2^{60}\)
\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)
\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\)
\(A=5\cdot\left(2+2^2+...+2^{58}\right)\)
Vậy A ⋮ 5
___________________
\(A=2+2^2+2^3+...+2^{60}\)
\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)
\(A=7\cdot\left(2+2^4+...+2^{58}\right)\)
Vậy A ⋮ 7
Bài 1:Chứng minh rằng :
a) 10^28+8 chia hết cho 72
b)8^8+2^20 chia hết cho17
Bài 2 :Cho :
a)A = 2+2^2+2^3+.........+2^60
chứng minh rằng Achia hết cho 3; 7; 15
a)$10^{28}$1028 chia 9 dư 1
8 chia 9 dư 8
1 + 8 = 9 chia hết cho 9
$\Rightarrow$⇒$10^{28}+8$1028+8 chia hết cho 9 (1)
$10^{28}$1028 chia hết cho 8 (vì có 3 chữ số tận cùng là 000 chia hết cho 8)
8 chia hết cho 8
$\Rightarrow$⇒$10^{28}+8$1028+8 chia hết cho 8 (2)
Từ (1) và (2) kết hợp với ƯCLN (8,9) = 1 . Suy ra $10^{28}+8$1028+8 chia hết cho 72
b)$8^8+2^{20}=\left(2^3\right)^8+2^{20}=2^{24}+2^{20}=2^{20}\times\left(2^4+1\right)=2^{20}\times17$88+220=(23)8+220=224+220=220×(24+1)=220×17 chia hết cho 17
cho a=2+22+23+...+260
chứng minh rằng achia hết cho 3 cho 7 và cho 15
A = 2 + 22 + 23 + .... + 260
= (2 + 22) + (23 + 24) + .... + (259 + 260)
= 2.(1 + 2) + 23.(1 + 2) + .... + 259.(1 + 2)
= 2.3 + 23.3 + .... + 259.3
= 3.(2 + 23 + ..... +259) chia hết cho 3
cho A= 1+3+32+33+..........+ 311 a. chứng minh rằng Achia hết cho 4 ;b.chứng minh rằng Achia hết 10;c.chứng minh rằng A chia hết cho 13
\(A=1+3+3^2+..........+3^{11}\)
\(\Leftrightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+.........+\left(3^{10}+3^{11}\right)\)
\(\Leftrightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+.........+3^{10}\left(1+3\right)\)
\(\Leftrightarrow A=1.4+3^2.4+.......+3^{10}.4\)
\(\Leftrightarrow A=4\left(1+3^2+..........+3^{10}\right)⋮4\left(đpcm\right)\)
A = 1 + 3 + 32 + 33 + ... + 311
A = ( 1 + 3 ) + ( 32 + 33 ) + ... + ( 310 + 311 )
A = 4 + 32 . ( 1 + 3 ) + ... + 310 . ( 1 + 3 )
A = 4 + 32 . 4 + ... + 310 . 4
A = 4 . ( 1 + 32 + ... + 310 ) \(⋮\) 4 ( Vì trong tích có một thừa số chia hết cho 4 )
~ Chúc bạn học giỏi ! ~
A = 1 + 3 + 32 + 33 + ... + 311
A = ( 1 + 3 + 32 ) + ... + ( 39 + 310 + 311 )
A = 13 + ... + 39 . ( 1 + 3 + 32 )
A = 13 + ... + 39 . 13
A = 13 . ( 1 + ... + 39 ) \(⋮\) 13 ( Vì trong tích có một thừa số chia hết cho 13 )
~ Chúc bạn học giỏi ! ~
Cho a = \(2+2^2+2^3+.....+2^{59}+2^{ }^{60}\)
a)Achia hết cho 2
B)a chia hết cho 3
C)a chia hết cho 7
hãy chứng minh a , b , c
a) \(A=2\left(1+2+2^2+...+2^{59}\right)⋮2\)
b) \(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
c) \(A=2\left(1+2+2^2\right)+2^5\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^5+...+2^{58}\right)⋮7\)
a) A = 2 + 2² + 2³ + ... + 2⁵⁹ + 2⁶⁰
= 2.(1 + 2 + 2² + ... + 2⁵⁸ + 2⁵⁹) 2
Vậy A ⋮ 2
b) A = 2 + 2² + 2³ + ... + 2⁵⁹ + 2⁶⁰
= (2 + 2²) + (2³ + 2⁴) + ... + (2⁵⁹ + 2⁶⁰)
= 2.(1 + 2) + 2³.(1 + 2) + ... + 2⁵⁹.(1 + 2)
= 2.3 + 2³.3 + ... + 2⁵⁹.3
= 3.(2 + 2³ + ... + 2⁵⁹) ⋮ 3
Vậy A ⋮ 3
c) A = 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + ... + 2⁵⁸ + 2⁵⁹ + 2⁶⁰
= (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2⁵⁸ + 2⁵⁹ + 2⁶⁰)
= 2.(1 + 2 + 2²) + 2⁴.(1 + 2 + 2²) + ... + 2⁵⁸.(1 + 2 + 2²)
= 2.7 + 2⁴.7 + ... + 2⁵⁸.7
= 7.(2 + 2⁴ + ... + 2⁵⁸) ⋮ 7
Vậy A ⋮ 7
cơ hội cho các bạn nhé.
Bài 1 : chứng minh rằng :
3^14 + 3^15 chia hết cho 4
A=1+3+3^2+3^3+...+3^2015
a, tính A
b, chứng minh Achia hết cho 4
Bài 1:Ta có:315+314=314.3+314=314.4 chia hết cho 4
Bài 2:a,\(3A=3+3^2+3^3+...........+3^{2016}\)
\(\Rightarrow3A-A=\left(3+3^2+.......+3^{2016}\right)-\left(1+3+.......+3^{2015}\right)\)
\(\Rightarrow2A=3^{2016}-1\Rightarrow A=\frac{3^{2016}-1}{2}\)
b,Ta có:A=1+3+32+33+.............+32015
=(1+3)+(32+33)+...............+(32014+32015)
=4+32.4+................+32014.4
=4.(1+32+.........+32014) chia hết cho 4
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cho a=2020+2020^2+2022^3+...+2020^2 chứng minh rằng Achia hết cho 2023
Biểu thức A viết có vẻ không đúng. Bạn xem lại đề.
Chứng minh rằng
a)14^14 - 1 chia hết cho 3
b)A=2+2^2+2^3+...+2^60 chia hết cho 15