Đáp án:
M=2+$2^{2}$ +$2^{3}$+............+$2^{19}$+$2^{20}$
M=2(1+2+2^2+2^3)+.....................+2^17(1+2+2^2+2^3)
M=2.15+..................+2^17.15
M=15(2+....+2^17) chia hết cho 15
⇒Mchia hết cho 5
b)ta có 2M=$2^{2}$ +$2^{3}$+............+$2^{19}$+$2^{20}$+$2^{21}$
lấy 2m-m=($2^{2}$ +$2^{3}$+............+$2^{19}$+$2^{20}$+$2^{21}$)-(2+$2^{2}$ +$2^{3}$+............+$2^{19}$+$2^{20}$)
M=$2^{21}$-2 <$2^{21}$
⇒M<$2^{21}$