Đáp án+ Giải thích các bước giải:
$a)2020^{21}+2020^{20}$ và $2021^{21}$
$\text{Ta có:$2020^{21}+2020^{20}=2020^{20}.(2020+1)=2020^{20}.2021$}$
$2021^{21}=2021^{20}.2021$
$\text{Vì 2020<2021⇒$2020^{20}.2021<2021^{20}.2021$}$
$\text{Hay $ 2020^{21}+2020^{20}<2021^{21}$}$
$b)A=3+2^{2}+2^{3}+2^{4}+....+2^{2001}+2^{2002}$ và $B=2^{2003}$
$\text{Ta có:$A=3+2^{2}+2^{3}+2^{4}+....+2^{2001}+2^{2002}$}$
$=>A=3+(2^{2}+2^{3}+2^{4}+....+2^{2001}+2^{2002})$
$\text{Đặt C=$2^{2}+2^{3}+2^{4}+....+2^{2001}+2^{2002}$}$
$=>2C=2^{3}+2^{4}+2^{5}+....+2^{2002}+2^{2003}$
$=>2C-C=2^{2003}-4$
$=>A=3+2^{2003}-4=2^{2003}-1<2^{2003}$
$\text{Vậy A<B}$
