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Sửa câu a
a)Ta có:
\(A=3+3^2+3^3+...+3^{99}\)
\(A=\left(3+3^2+3^3\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\)
\(A=\left(3+3^2+3^3\right)+...+3^{96}.\left(3+3^2+3^3\right)\)
\(A=39+...+3^{96}.39\)
\(A=39.\left(1+...+3^{96}\right)\)
Vì 39 \(⋮\) 13 nên 39 . ( 1 + ... + 396 ) \(⋮\) 13
Vậy A \(⋮\) 13
_________
b)Ta có:
\(B=5+5^2+5^3+...+5^{50}\)
\(B=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{49}+5^{50}\right)\)
\(B=\left(5+5^2\right)+5^2.\left(5+5^2\right)+...+5^{48}.\left(5+5^2\right)\)
\(B=30+5^2.30+...+5^{48}.30\)
\(B=30.\left(1+5^2+...+5^{48}\right)\)
Vì 30 \(⋮\) 6 nên 30. ( 1 + 52 + ... + 548 ) \(⋮\) 6
Vậy B \(⋮\) 6
a,A=3+32+33+..+399=(3+32+33)+...+(397+398+399)
=3(1+3+32)+...+397(1+3+32)=3x13+...+397x13=13(3+...+97)⋮13
b,B=5+52+...+550=(5+52)+...+(549+550)=5(1+5)+..+549(1+5)
=5x6+...+549x6=6(5+..+549)⋮6.
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
a: \(B=3^1+3^2+...+3^{2010}\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4\left(3+3^3+...+3^{2009}\right)⋮4\)
\(B=3\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2008}\right)⋮13\)
b: \(C=5^1+5^2+...+5^{2010}\)
\(=5\left(1+5\right)+...+5^{2009}\left(1+5\right)\)
\(=6\left(5+...+5^{2009}\right)⋮6\)
\(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)\)
\(=31\left(5+...+5^{2008}\right)⋮31\)
c: \(D=7\left(1+7\right)+...+7^{2009}\left(1+7\right)\)
\(=8\left(7+...+7^{2009}\right)⋮8\)
\(D=7\left(1+7+7^2\right)+...+7^{2008}\left(1+7+7^2\right)\)
\(=57\left(7+...+7^{2008}\right)⋮57\)
cho C=5+52+53+54+...+520 chứng minh rằng:
a)C chia hết cho 5 b) C chia hết cho 6 c) C chia hết cho 13
\(a,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)
\(=5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)\)
Ta thấy: \(5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)⋮5\)
nên \(C⋮5\)
\(b,C=5+5^2+5^3+5^4\cdot\cdot\cdot+5^{20}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdot\cdot\cdot+\left(5^{19}+5^{20}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdot\cdot\cdot+5^{19}\left(1+5\right)\)
\(=5\cdot6+5^3\cdot6+\cdot\cdot\cdot+5^{19}\cdot6\)
\(=6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)\)
Ta thấy: \(6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)⋮6\)
nên \(C⋮6\)
\(c,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)
\(=\left(5+5^3\right)+\left(5^2+5^4\right)+\cdot\cdot\cdot+\left(5^{17}+5^{19}\right)+\left(5^{18}+5^{20}\right)\)
\(=5\left(1+5^2\right)+5^2\left(1+5^2\right)+\cdot\cdot\cdot+5^{17}\cdot\left(1+5^2\right)+5^{18}\left(1+5^2\right)\)
\(=5\cdot26+5^2\cdot26+\cdot\cdot\cdot+5^{17}\cdot26+5^{18}\cdot26\)
\(=26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)\)
Ta thấy: \(26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)⋮13\)
nên \(C⋮13\)
#\(Toru\)
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 + ... + 5^19 )
=> C chia hết cho 5
b,
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 ) + 5^3 . ( 1 + 5 ) + ... + 5^19 . ( 1 + 5 )
=> C = 5 . 6 + 5^3 . 6 + ... + 5^19 . 6
=> C = 6 . ( 5 + 5^3 + ... + 5^19 )
=> C chia hết cho 6
c,
C = 5 + 5^2 + 5^3 + ... + 5^20
=> C = (5 + 5^2 + 5^3 + 5^4 ) + ... + (5^17 + 5^18 + 5^19 + 5^20 )
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 ) + ... + 5^17 . ( 1+ 5 + 5^2 +5^3)
=> C = 5 . 156 + 5^5 . 156 + ...+ 5^17 . 156
=> C = 5 . 12 . 13 + 5^5 . 12 . 13 + ... + 5^17 . 12 . 13
=> C = 13 . ( 5 . 12 + 5^5 . 12 + ... + 5^17 . 12 )
=> C chia hết cho 13
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
\(A=3+3^2+3^3+...+3^{99}\\ \Rightarrow A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\\ \Rightarrow A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{97}\left(1+3+3^2\right)\\ \Rightarrow A=\left(1+3+3^2\right)\left(3+3^4+...+3^{97}\right)\\ \Rightarrow A=13\left(3+3^4+...+3^{97}\right)⋮13\)
\(A=3+3^2+3^3+...+3^{99}\\ 3A-A=3^{99}-1\\ A=\dfrac{3^{99}-1}{2}\)
Đặt A = 3² + 3³ + 3⁴ + ... + 3⁹⁹
= 3² + 3³ + (3⁴ + 3⁵ + 3⁶) + (3⁷ + 3⁸ + 3⁹) + ... + (3⁹⁷ + 3⁹⁸ + 3⁹⁹)
= 36 + 3⁴.(1 + 3 + 3²) + 3⁷.(1 + 3 + 3²) + ... + 3⁹⁷.(1 + 3 + 3²)
= 36 + 3⁴.13 + 3⁷.13 + ... + 3⁹⁷.13
= 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷)
Do 36 không chia hết cho 13
13.(3⁴ + 3⁷ + ... + 3⁹⁷) ⋮ 13
⇒ 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷) không chia hết cho 13
⇒ A không chia hết cho 13
Em xem lại đề nhé, có thể em viết thiếu số 3 rồi
A = 3+32+33+.....+399
A = 3.(1+3+9) + 34.( 3+1+9) + 37.( 3+1+9) + .... + 397.(1+3+9)
A = 3.13 + 34.13 + 37.13+....+397.13
Vậy A chi hết cho 13
B = 5+52+53+54+...+550
B = 5.(5+1) + 53.(5+1) + ..... + 559.(1+5)
B = 5.6 + 53.6 + .... + 559.6
Vậy B chia hết cho 6
A