Đáp án:
${2^{2021}} $
Lời giải:
$\begin{array}{l}
A = {2^2} + {2^3} + {2^4} + {2^5} + ... + {2^{2020}}\\
\Rightarrow 2.A = 2.\left( {{2^2} + {2^3} + {2^4} + {2^5} + ... + {2^{2020}}} \right)\\
\Rightarrow 2.A = {2^3} + {2^4} + {2^5} + {2^6} + ... + {2^{2021}}\\
\Rightarrow 2.A - A = {2^3} + {2^4} + {2^5} + {2^6} + ... + {2^{2021}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{2^2} + {2^3} + {2^4} + {2^5} + ... + {2^{2020}}} \right)\\
\Rightarrow A = {2^3} + {2^4} + {2^5} + {2^6} + ... + {2^{2021}} - {2^2} - {2^3} - {2^4} - {2^5} - ... - {2^{2020}}\\
\Rightarrow A = {2^{2021}} - {2^2}
\end{array}$
Do đó:
$2^2+ {2^2} + {2^3} + {2^4} + {2^5} + ... + {2^{2020}}$
$=2^2+A$
$=2^2+{2^{2021}} - {2^2}$
$={2^{2021}} $
