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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:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
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: \(D=7\left(1+7\right)+...+7^{2009}\left(1+7\right)\)
\(=8\left(7+...+7^{2009}\right)⋮8\)
\(D=7\left(1+7+7^2\right)+...+7^{2008}\left(1+7+7^2\right)\)
\(=57\left(7+...+7^{2008}\right)⋮57\)
\(B=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{58}\right)⋮7\)
Ta có: 52003 + 52002 + 52001
= 52001.(52 + 5 + 1)
= 52001 . 31 chia hết cho 31
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
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.
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 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=1+3+32+33+...+32000 ⋮ 13
=(1+3+32)+(33+34+35)+...+(31998+31999+32000) ⋮ 13
=1.(1+3+32)+33.(1+3+32)+...+31998.(1+3+32) ⋮ 13
=1.13+33.13+...+31998.13 ⋮ 13
=13.(1+33+...+31998) ⋮ 13
Vì 13 ⋮ 13 nên 1+3+32+33+...+32000 ⋮ 13
B=1+7+72+73+...+72018 ⋮ 19
=(1+7+72)+(73+74+75)+...+(72016+72017+72018) ⋮ 19
=1.(1+7+72)+73.(1+7+72)+...+72016.(1+7+72) ⋮ 19
=1.57+73.57+...+72016.57 ⋮ 19
=57.(1+73+...+72016) ⋮ 19
Vì 57 ⋮ 19 nên 1+7+72+73+...+72018 ⋮ 19
\(A=1+3+3^2+...+3^{2000}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{1998}+3^{1999}+3^{2000}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1998}\left(1+3+3^2\right)\)
\(=13+13.3^3+...+13.3^{1998}\)
\(=13.\left(1+3^3+...+3^{1998}\right)⋮13\)
\(B=1+7+7^2+\left(7^3+7^4+7^5\right)+...+\left(7^{2016}+7^{2017}+7^{2018}\right)\)
\(=1+7+7^2+7^3\left(1+7+7^2\right)+...+7^{2016}\left(1+7+7^2\right)\)
\(=57+57.7^3+...+57.7^{2016}\)
\(=57.\left(1+7^3+...+7^{2016}\right)\)
Mà \(57⋮19\Rightarrow B⋮19\)
Câu C làm tương tự 2 câu trên (vẫn tách nhóm 3 số hạng)