B=4+32+33+.....+32003
<=> B=4+(32+33+.....+32003)
Đặt A=32+33+......+32003
=> 3A=3(32+33+....+32003)
=> 3A=33+34+....+32004
=> 3A-A=(33+34+....+32004)-(32+33+....+32003)
=> 2A=32004-32
=> \(A=\frac{3^{2004}-3^2}{2}\)
Thay \(A=\frac{3^{2004}-3^2}{2}\)và B, ta có:
\(B=2^2+\frac{3^{2004}-3^2}{2}\)
=> B<C