A=3+3^2+3^3+...+3^2004
Ta có:A=(3+3^2+3^3+3^4)+...+(3^2001+3^2002+3^2003+3^2004)
=>A=120+...+(3^2000.3+3^2000.3^2+3^2000.3^3+3^2000.3^4)
=>A=120+...+3^2000(3+3^2+3^3+3^4)
=>A=120+...+3^2000.120
=>A=(1+...+3^2000).120
Vì 120 chia hết cho 120 nên A chia hết cho 120=>A chia hết cho 10
A=3+3^2+3^3+...+3^2004
=>A=(3+3^2+3^3)+...+(3^2002+3^2003+3^2004)
=>A=39+...+(3^2000.3+3^2000.3^2+3^2000.3^3)
=>A=39+...+3^2000(3+3^2+3^3)
=>A=39+...+3^2000.39
=>A=(1+...+3^2000).39
Vì 39 chia hết cho 13 nên A chia hết cho 13
Ta có:A chia hết cho 10;A chia hết cho 13 và (10;13)=1 nên A chia hết cho 10.13
=>A chia hết cho 130
Vậy...
Đặt A=1.2+2.3+3.4+............+1999.2000
3A=1.2.3+2.3.(4-1)+.................+1999.2000.(2001-1998)
3A=1.2.3+2.3.4-1.2.3+............+1999.2000.2001-1998.1999.2000
3A=1999.2000.2001
A=1999.2000.2001:3
A=2666666000
b,Đặt B=1.2+2.2+3.3+............+1999.1999
B=1.(2-1)+2.(3-1)+3.(4-1)+..........+1999.(2000-1)
B=1.2-1+2.3-2+3.4-3+...........+1999.2000-1999
B=(1.2+2.3+3.4+.............+1999.2000)-(1+2+3+...........+1999)
B=2666666000-1999000
B=2664667000
c,Đặt C=1.2.3+2.3.4+..........+48.49.50
4C=1.2.3.4+2.3.4.(5-1)+.........+48.49.50.(51-47)
4C=1.2.3.4+2.3.4.5-1.2.3.4+..............+48.49.50.51-47.48.49.50
4C=48.49.50.51
C=48.49.50.51:4
C=1499400