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A = (3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+.....+(3^97+3^98+3^99+3^100)
= 120+3^4.(3+3^2+3^3+3^4)+.....+3^96.(3+3^2+3^3+3^4)
= 120+3^4.110+....+3^96.120
= 120.(1+3^4+.....+3^96) chia hết cho 120
=> ĐPCM
Tk mk nha
ta co A=(31+32+33+34)+...+(397+398+399+3100)
tớ gợi ý nhiêu đây thôi
\(A=2+2^2+2^3+...+2^{100}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(A=6+2^2.6+...+2^{98}.6\)
\(A=6\left(1+2^2+...+2^{98}\right)\)
Có : \(6⋮6\)
\(\Rightarrow A=6\left(1+2^2+...+2^{98}\right)⋮6\)
\(\Rightarrow A⋮6\)
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a = 2 + 22 +23+........................+ 2100 chia hết cho 62
a = [ 2 + 22 +23+.24+25 ] +[ 26 +27 +28+29+210 ] + ...........+ [ 296 + 297 +298 +299 + 2100 ]
a= 62 + [ 210 . 62 ] + [ 215 . 62 ] + [ 220. 62 ] + ......................+ [ 2100 . 62 ]
a= 62 . [ 210 + 215 + 220 +......................+ 2100 ]
Mà 62 chia hết cho 62 => 62 . [ 210 + 215 + 220 +......................+ 2100 ] hay a chia hết cho 62
a = (2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)+.....+(2^96+2^97+2^98+2^99+2^100)
= 62+2^5.(2+2^2+2^3+2^4+2^5)+......+2^95.(2+2^2+2^3+2^4+2^5)
= 62+2^5.62+....+2^95.62
= 62.(1+2^5+....+2^95) chia hết cho 62
=> ĐPCM
k mk nha
S=1+7+7^2+7^3+...+7^100+7^101
=(1+7)+7^2(1+7)+...+7^100(1+7)
=8+7^2.8+...+7^100.8
=8.(1+7^2+...+7^100) chia hết cho 8
Vậy S chia hết cho 8
a.S=4+4^2+4^3+4^4+...+4^99+4^100 chia hết cho 5
S=(4+4^2)+(4^3+4^4)+...+(4^99+4^100)
S=20+4^2*20+...+4^98
S=20*(1+4^2+...+4^98) chia hết cho 5(đpcm)
b.S=2+2^2+2^3+2^4+...+2^2009+2^2010CHIA HẾT CHO 6
S=(2+2^2)+(2^3+2^4)+...+(2^2009+2^2010)
S=6+2^2.*6+...+2^2008
S=6*(1+2^2+...+2^2008)CHIA HẾT CHO 6
\(a=2+2^2+2^3+2^4+...+2^{100}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{99}\right)⋮3\).
Sửa đề: \(A=2^0+2^1+2^2+...+2^{99}\)
\(=\left(2^0+2^1\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)
\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{98}\left(1+2\right)\)
\(=3\left(1+2^2+...+2^{98}\right)⋮3\)
Ta có:
A = 4 + 42 + 43 + 44 + ... + 499 + 4100
A = (4 + 42) + (43 + 44) + ... + (499 + 4100)
A = 4(1 + 4) + 43(1 + 4) + ... + 499(1 + 4)
A = 4.5 + 43.5 + ... + 499.5
A = 5.(4 + 43 + ... + 499)
Vậy A chia hết cho 5
\(a,A=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)...+\left(2^{99}+2^{100}\right)\)
\(=6+2^2\cdot\left(2+2^2\right)+2^4\cdot\left(2+2^2\right)...+2^{98}\cdot\left(2+2^2\right)\)
\(=6+2^2\cdot6+2^4\cdot6...+2^{98}\cdot6\)
\(=6\cdot\left(1+2^2+2^4+...+2^{98}\right)\)
Vì \(6\cdot\left(1+2^2+2^4+...+2^{98}\right)⋮6\)
nên \(A⋮6\)
\(b,A=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^3\right)+\left(2^2+2^4\right)+\left(2^3+2^5\right)+...+\left(2^{97}+2^{99}\right)+\left(2^{98}+2^{100}\right)\)
\(=10+2\cdot\left(2+2^3\right)+2^2\cdot\left(2+2^3\right)+...+2^{96}\cdot\left(2+2^3\right)+2^{97}\cdot\left(2+2^3\right)\)
\(=10+2\cdot10+2^2\cdot10+...+2^{96}\cdot10+2^{97}\cdot10\)
\(=10\cdot\left(1+2+2^2+...+2^{96}+2^{97}\right)\)
Vì \(10\cdot\left(1+2+2^2+...+2^{96}+2^{97}\right)⋮10\)
nên \(A⋮10\)
#\(Toru\)
Ta có: A = 2 + 22 + 23 + 24 + ... + 299 + 2100
A = (2 + 22) + (23 + 24) + ... + (299 + 2100)
A = 6 + 22(2 + 22) + .... + 298(2 + 22)
A = 6 + 22.6 + ... + 298.6
A = 6.(1 + 22 + ... + 298) \(⋮\)6
cho 31