ta có : \(B=2004+2004^2+2004^3+...+2004^{10}\)

\(B=\left(2004+2004^2\right)+\left(2004^3+2004^4\right)+...+\left(2004^9+2004^{10}\right)\)

\(B=2004.\left(1+2004\right)+2004^3\left(1+2004\right)+...+2004^9\left(1+2004\right)\)

\(B=2004.2005+2004^3.2005+...+2004^9.2005\)

\(B=2005.\left(2004+2004^3+...+2004^9\right)⋮2005\)

\(\Rightarrow2005.\left(2004+2004^3+2004^9\right)\) chia hết cho \(2005\)

\(\Leftrightarrow B=2004+2004^2+2004^3+...+2004^{10}\) chia hết cho \(2005\) (đpcm)

(đpcm)

C = 2004 + 2004²+2004³+2004⁴+...+2004¹⁰

C = (2004 +2004²)+(2004³+2004⁴)+...+(2004⁹+2004¹⁰)

C = 2004(1+2004) + 2004³ .(1+2004)+...+ 2004⁹. (1+2004)

C = 2004 .2005 + 2004³ .2005+....+2004⁹.2005

C = 2005.(2004+2004³ + ...+2004⁹)

Vì 2005 chia hết cho 2005 => 2005.(2004+2004³ + ...+2004⁹) chia hết cho 2005 => C chia hết cho 2005(ĐPCM)

Ta có : 

\(C=2004+2004^2+2004^3+...+2004^9+2004^{10}\)

\(=\left(2004+2004^2\right)+\left(2004^3+2004^4\right)+...+\left(2004^9+2004^{10}\right)\)

\(=2004\left(1+2004\right)+2004^3\left(1+2004\right)+...+2004^9\left(1+2004\right)\)

\(=2004.2005+2004^3.2005+...+2004^9.2005\)

\(=2005\left(2004+2004^3+...+2004^9\right)⋮2005\left(đpcm\right)\)

Ngô Thị Diệu Linh

C = 2004 + 20042+20043+20044+...+200410
C = (2004 +20042)+(20043+20044)+...+(20049+200410)

C = 2004(1+2004) + 20043 .(1+2004)+...+ 20049. (1+2004)

C = 2004 .2005 + 20043 .2005+....+20049.2005

C = 2005.(2004+20043 + ...+20049)

Vì 2005 chia hết cho 2005 => 2005.(2004+20043 + ...+20049) chia hết cho 2005 => C chia hết cho 2005(ĐPCM)

ta có A=2004+2004²+...........................................+2004¹⁰                                                                                                                                    tương đương A=2004.(1+2004)+2004².(1+2004)+..............+2004⁹(1+2004)

                           A=2004.2005+2004².2005......................+2004⁹.2005

             ta có    A=2005(2004+2004²................2004⁹)

                 suy ra A=[ 2005(2004+2004²...............2004⁹)] chia hết cho 2005

                tương đưong A=(2004+2004²................+2004¹⁰) chia hết cho 2005

tham khảo ở link: https://olm.vn/hoi-dap/detail/87851120650.html

Đặt A=2004+2004²+2004³+...+2004¹⁰

         =(2004+2004²)+(2004³+2004⁴)+...+(2004⁹+2004¹⁰)

         =2004(1+2004)+2004³(1+2004)+...+2004⁹(1+2004)

         =2004.2005+2004³.2005+...+2004⁹.2005 chia hết cho 2005

Vậy A chia hết cho 2005.

Đặt A=2004+2004²+2004³+...+3004¹⁰

    A=(2004+2004²)+(2004³+2004⁴)+.....+(2004⁹+2004¹⁰⁾

      A=(2004+2004²)+2004².(2004+2004²)+...+2004⁸.(2004+2004²)

    A=4018020+2004².4018020+.....+2004⁸.4018020

     A=4018020.(1+2004²+....+2004⁸)

     A=2004.2005.(1+2004²+...+2004⁸)\(⋮\)2005

Vậy A\(⋮\)2005

Chúc bn học tốt

A =   (2004 + 2004² ) + ( 2004³ + 2004⁴)+  (2004⁵ + 2004⁶)  +............................+ (2004⁸ + 2004¹⁰)
A = 2004 ( 1 + 2004 ) + 2004³ ( 1 +2004 ) + .... + 2004⁸ ( 1+ 2004 )
A = 2004.2005 + 2004³.2005 +....+2004⁸.2005
A = 2005.(  2004 + 2004² + 2004³ + 2004⁴ +  2004⁵ + 2004⁶  +............................+ 2004⁸ + 2004¹⁰   )
Vậy A chia hết cho 2005

có sai đề ở chỗ 2004^8+2004^10 ko bn

a) 5+5²+5³+5⁴+...+5¹⁰⁰

= (5+5²)+(5³+5⁴)+...+(5⁹⁹+5¹⁰⁰)

= 30+5².(5+5²)+...+5⁹⁸.(5+5²)

= 30+5².30+...+5⁹⁸.30

= 30.(1+5²+...+5⁹⁸)

Vì 30 chia hết cho 10

=> 30.(1+5²+...+5⁹⁸) chia hết cho 10

=> 5+5²+5³+...+5¹⁰⁰ chia hết cho 10