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\(10^6\) tận cùng là 0 \(=>10^6+2\) tận cùng là 2 \(=>10^6+2\) chia hết cho 2
a=2+2^2+2^3+...+2^10
a=(2+2^2)+(2^3+2^4)+...+(2^9+2^10)
a=2.(1+2)+2^3.(1+2)+...+2^9.(1+2)
a=3.(2+2^3+...+2^9)
=> a chia hết cho 3
a=2+2^2+2^3+...+2^10
a=(2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)
a=2.(1+2+4+8+16)+2^6.(1+2+4+8+16)
a=31.(2+2^6)
=> a chia hết cho 31
chúc bạn học tốt nha
1+7+7 mũ 2+7 mũ 3......+7 mũ 100.Tính a,a là tổng dãy số trên
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
\(A=2+2^2+2^3+.......+2^{100},\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+....+\left(2^{99}+2^{100}\right)\)
\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+.....+2^{98}\left(2+2^2\right)\)
\(A=6+2^2.6+....+2^{98}.6\)
\(A=6\left(1+2^2+.......+2^{98}\right)\)
\(A=6\left(1+2^2+........+2^{98}\right)\text{⋮6}\)
Gọi C là giá trị của biểu thức trên
a) CMR : C chia hết cho 31
\(C=2+2^2+2^3+...+2^{99}+2^{100}\)
\(C=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{19}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(C=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(C=2.31+2^6.31+...+2^{96}.31\)
\(C=31\left(2+2^6+2^{10}+...+2^{96}\right)⋮31\)(đpcm)
b) CMR : C chia hết cho 5
\(C=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)
\(=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{97}+2^{99}\right)+\left(2^{98}+2^{100}\right)\)
\(=2\left(1+2^2\right)+2^2\left(1+2^2\right)+...+2^{97}\left(1+2^2\right)+2^{98}\left(1+2^2\right)\)
=\(2.5+2^2.5+...+2^{97}.5+2^{98}.5\)
\(=5\left(2+2^2+...+2^{97}+2^{98}\right)⋮5\)(đpcm)
Vậy 2 + 2^2 + 2^3 + ...+ 2^98 + 2^99 + 2^100 vừa chia hết cho 5 vừa chia hết cho 31
a) CMR : C chia hết cho 31
\(C = 2 + 2^{2} + 2^{3} + . . . + 2^{99} + 2^{100}\)
\(C = \left(\right. 2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} \left.\right) + \left(\right. 2^{6} + 2^{7} + 2^{8} + 2^{9} + 2^{19} \left.\right) + . . . + \left(\right. 2^{96} + 2^{97} + 2^{98} + 2^{99} + 2^{100} \left.\right)\)
\(C = 2 \left(\right. 1 + 2 + 2^{2} + 2^{3} + 2^{4} \left.\right) + 2^{6} \left(\right. 1 + 2 + 2^{2} + 2^{3} + 2^{4} \left.\right) + . . . + 2^{96} \left(\right. 1 + 2 + 2^{2} + 2^{3} + 2^{4} \left.\right)\)
\(C = 2.31 + 2^{6} . 31 + . . . + 2^{96} . 31\)
\(C = 31 \left(\right. 2 + 2^{6} + 2^{10} + . . . + 2^{96} \left.\right) 31\)(đpcm)
b) CMR : C chia hết cho 5
\(C = 2 + 2^{2} + 2^{3} + 2^{4} + . . . + 2^{97} + 2^{98} + 2^{99} + 2^{100}\)
\(= \left(\right. 2 + 2^{3} \left.\right) + \left(\right. 2^{2} + 2^{4} \left.\right) + . . . + \left(\right. 2^{97} + 2^{99} \left.\right) + \left(\right. 2^{98} + 2^{100} \left.\right)\)
\(= 2 \left(\right. 1 + 2^{2} \left.\right) + 2^{2} \left(\right. 1 + 2^{2} \left.\right) + . . . + 2^{97} \left(\right. 1 + 2^{2} \left.\right) + 2^{98} \left(\right. 1 + 2^{2} \left.\right)\)
=\(2.5 + 2^{2} . 5 + . . . + 2^{97} . 5 + 2^{98} . 5\)
\(= 5 \left(\right. 2 + 2^{2} + . . . + 2^{97} + 2^{98} \left.\right) 5\)(đpcm)
Vậy 2 + 2^2 + 2^3 + ...+ 2^98 + 2^99 + 2^100 vừa chia hết cho 5 vừa chia hết cho 31