a, Ta có:

2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100

=  2 + 2 2 + 2 3 + 2 4 + 2 5 +...+ 2 96 + 2 97 + 2 98 + 2 99 + 2 100

= 2. 1 + 2 + 2 2 + 2 3 + 2 4 +...+ 2 96 1 + 2 + 2 2 + 2 3 + 2 4

=  2 . 31 + 2 6 . 31 + . . . + 2 96 . 31

=  2 + 2 6 + . . . + 2 96 . 31  chia hết cho 31

b, Ta có:

5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150

=  5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150

=  5 1 + 5 + 5 3 1 + 5 + 5 5 1 + 5 + . . . + 5 149 1 + 5

=  5 . 6 + 5 3 . 6 + 5 5 . 6 + . . . + 5 149 . 6

=  ( 5 + 5 3 + 5 5 + . . . + 5 149 ) . 6  chia hết cho 6

Ta lại có:

5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150

=  5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 +...+ 5 145 + 5 146 + 5 147 + 5 148 + 5 149 + 5 150  (có đúng 25 nhóm)

=  [ ( 5 + 5 4 ) + ( 5 2 + 5 5 ) + ( 5 3 + 5 6 ) ] + ... +  [ 5 145 + 5 148 ) + ( 5 146 + 5 149 ) + ( 5 147 + 5 150 ]

=  [ 5 ( 1 + 5 3 ) + 5 2 ( 1 + 5 3 ) + 5 3 ( 1 + 5 3 ) ] + ... +  [ 5 145 1 + 5 3 ) + 5 146 ( 1 + 5 3 ) + 5 147 ( 1 + 5 3 ]

=  ( 5 . 126 + 5 2 . 126 + 5 3 . 126 )...

Chịu 🤭🤭🤭

Sửa đề: \(B=2+2^2+2^3+...+2^{100}\)

\(=2\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{97}\right)⋮5\)

\(B=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

\(=31\left(2+2^6+...+2^{96}\right)⋮31\)

did you studied at le van tam primary school

giúp với mình đang cần gấp

\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+...+3^{58}\right)\\ A=13\left(3+...+3^{58}\right)⋮13\)

\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\\ M=\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\\ M=\left(2+2^2+2^3+2^4\right)\left(1+...+2^{16}\right)\\ M=30\left(1+...+2^{16}\right)⋮5\)

a: \(A=3+3^2+3^3+\cdots+3^{60}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{58}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+\cdots+3^{58}\right)\)

=>A⋮13

b: \(M=2+2^2+\cdots+2^{20}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{17}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+\cdots+2^{17}\right)\)

=>M⋮5

a: \(A=3+3^2+3^3+\cdots+3^{60}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{58}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+\cdots+3^{58}\right)\)

=>A⋮13

b: \(M=2+2^2+\cdots+2^{20}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{17}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+\cdots+2^{17}\right)\)

=>M⋮5

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