Đáp án:
$${2^{32}} - 1$$
Giải thích các bước giải:
$$\eqalign{
& \left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr
& = \left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr
& = \left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr
& = \left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right) \cr
& = \left( {{2^{16}} - 1} \right)\left( {{2^{16}} + 1} \right) = {2^{32}} - 1 \cr} $$
