c1:ta có S=2+2^2+2^3+...+2^99+2^100 
=>nhóm 5 số đầu lấy 2 ra ngoài ta sẽ được2 nhân với 31 
tương tự với các số sau 
có số số hạng của S là 100 chia hết cho 5 nên ta sẽ được 20 cặp có nhân tử là 31 cuốicùng đặt 31 ra ngoài làm nhân tử chung thì được dpcm 
c2:S = 2 + 2^2+2^3+...+2^99+2^100 
Suy ra 2S = 2^2 + 2^3 + 2^4 + ... + 2^100 + 2^101 
2S - S = 2^2 + 2^3 + 2^4 + ... + 2^100 + 2^101 - (2 + 2^2 + 2^3 + ... + 2^99 + 2^100) = 2^101 - 2 
S = 2^101 - 2 = 2 (2^100 - 1) 
2^5 = 32 đồng dư với 1 modun 31 
Suy ra (2^5)^20 đồng dư với 1 modun 31 
Hay 2^100 đồng dư với 1 modun 31 
Nên 2^100 - 1 chia hết cho 31 
Vậy S chia hết cho 31

 :

\(A = 2 + 2^{2} + 2^{3} + 2^{4} + . . . + 2^{99} + 2^{100}\)

=> \(A = \left(\right. 2 + 2^{2} \left.\right) + \left(\right. 2^{3} + 2^{4} \left.\right) + . . . + \left(\right. 2^{99} + 2^{100} \left.\right)\)

=> \(A = 2 \left(\right. 1 + 2 \left.\right) + 2^{3} \left(\right. 1 + 2 \left.\right) + . . . + 2^{99} \left(\right. 1 + 2 \left.\right)\)

=> \(A = 2.3 + 2^{3} . 3 + . . . + 2^{99} . 3\)

=> \(A = \left(\right. 2 + 2^{3} + . . . + 2^{99} \left.\right) . 3\)chia hết cho 3             ( 1 )

Ta lại có :

\(A = 2 + 2^{2} + 2^{3} + 2^{4} + . . . + 2^{99} + 2^{100}\)

=> \(A = 2 \left(\right. 1 + 2 + 2^{2} + 2^{3} + . . . + 2^{98} + 2^{99} \left.\right)\)chia hết cho 2       ( 2 )

Từ ( 1 ) và ( 2 ) ta có :

A chia hết cho 2 . 3 hay A chia hết cho 6

Ta có :

\(A = 2 + 2^{2} + 2^{3} + 2^{4} + . . . + 2^{99} + 2^{100}\)

=> ​\(A = \left(\right. 2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} \left.\right) + . . . . \left(\right. 2^{96} + 2^{97} + 2^{98} + 2^{99} + 2^{100} \left.\right)\)​

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

=> \(A = 2.31 + . . . + 2^{96} . 31\)

=> \(A = \left(\right. 2 + . . . + 2^{96} \left.\right) 31\)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

b,Ta có :

C=2+2^2+2^3+2^4+2^5...+2^96+2^97+2^98+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+2.2^2+2.2^3+2.2^4)+...+(2^96.1+2^96.2+2^96.2^2+2^96.2^3+2^96.2^4)

  =2. (1+2+2^2+2^3+2^4) +...+2^96.(1+2+2^2+2^3+2^4)

  =2.31+...+2^96.31

  =31. (2+...+2^96) chia hết cho 31

=>C chia hết cho 31

1. Chứng minh rằng: 3^2+3^3+3^4+...+3^101 chia hết cho 120.

Ta có:

A=3^2+3^3+3^4+...+3^101 

= (3^2+3^3+3^4+3^5) + ( 3^6+3^7+3^8+3^9) +.... + ( 3^98 + 3^99 + 3^100 + 3^101)

= 3.(3+3^2+3^3+3^4) + 3^5.(3+3^2+3^3+3^4) +....+ 3^97.(3+3^2+3^3+3^4)

= 120.(3+3^5+...+3^97) chia hết cho 120

\(S = 1 + 2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} + 2^{7}\)

\(\Rightarrow S = \left(\right. 1 + 2 \left.\right) + \left(\right. 2^{2} + 2^{3} \left.\right) + \left(\right. 2^{4} + 2^{5} \left.\right) + \left(\right. 2^{6} + 2^{7} \left.\right)\)

\(\Rightarrow S = \left(\right. 1 + 2 \left.\right) + 2^{2} \left(\right. 1 + 2 \left.\right) + 2^{4} \left(\right. 1 + 2 \left.\right) + 2^{6} \left(\right. 1 + 2 \left.\right)\)

\(\Rightarrow S = \left(\right. 1 + 2 \left.\right) \left(\right. 1 + 2^{2} + 2^{4} + 2^{6} \left.\right)\)

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

a) S = 2 + 2² + 2³ + 2⁴ +.....+ 2⁹ + 2¹⁰

   = (2 + 2²) + (2³ + 2⁴) +.....+ (2⁹ + 2¹⁰)

   = 2(1 + 2) + 2³(1 + 2) +....+ 2⁹(1 + 2)

   = 3.(2 + 2³ +.... + 2⁹) chia hết cho 3

   => S = 2 + 2² + 2³ + 2⁴ +.....+ 2⁹ + 2¹⁰ chia hết cho 3 (Đpcm)

b) 1+3²+3³+3⁴+...+3⁹⁹

=(1+3+3²+3³)+....+(3⁹⁶+3⁹⁷+3⁹⁸+3⁹⁹)

=40+.........+3⁹⁶.40

=40.(1+....+3⁹⁶) chia hết cho 40 (đpcm)

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