Đặt :
\(B=2.1+2.3+2.3^2+....+2.3^{2004}\)
\(\Rightarrow B=2\left(1+3+3^2+....+2^{2004}\right)\)
\(\Rightarrow B=2\left[\left(1+3^2\right)+3\left(1+3^2\right)+....+3^{2002}\left(1+3^2\right)\right]\)
\(\Rightarrow B=2\left[10+3.10+....+3^{2002}.10\right]\)
\(\Rightarrow B=20.\left[1+3+....+3^{2002}\right]\)
Vậy B có tận cùng là 0
Đặt C=2.1+2.3+...+2.32004
C=2.(1+3+...+32004)
đặt D=1+3+...+32004
3D=3+32+...+32005
D=(32005-1):2
D=[3 . (34)501-1]:2
D=[3. 81501-1]:2
D=[3 x ...1-1]:2
D=[...3-1]:2
D=...2:2
D=...1
vậy C=2.D
C=2 x ...1
C=...2