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 : \(6⋮6\)

\(\Rightarrow A=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(\Rightarrow A⋮6\)

suuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

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

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

\(\)A = (2 + 2\(^2\)) + 2\(^2\)(2 + 2\(^2\)) +....+ 2\(^{98}\) (2 + 2\(^2\))

A = 6 + 2\(^2\).6 + .... + 2\(^{98}\).6

A = 6 (1 + 2\(^2\) + .... + 2\(^{98}\)) \(⋮\) 6

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

haha mình đang cần câu trả lời

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)Cm A=10mũ99 cộng 104 chia hết cho hai và ba 

b)Cm B=10 mũ 100 cộng 17 chia hết cho 9

c)Cm 10 mũ 11 cộng với 8 chia hết cho 18 với n thuộc z và n bé hơn hoặc bằng 2

mong mọi người trả lời giúp mik cảm ơn các bạn

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

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

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

\(=3\left(2+2^3+...+2^{99}\right)⋮3\).

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

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

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

\(A=4.\left(1+4\right)+4^3.\left(1+4\right)+...+4^{99}.\left(1+4\right)\)

\(A=4.5+4^3.5+..4^{99}.5\)

\(A=5.\left(4+4^3+...4^{99}\right)\)

\(\Rightarrow A⋮5\)

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)

=> A chia hết cho 5 (đpcm)