Giải thích các bước giải:

Bài 1:

Số số hạng của tổng đã cho là: \(\left( {90 - 1} \right):1 + 1 = 90\). Do đó, ta có:

\(\begin{array}{l}
*)\\
A = 4 + {4^2} + {4^3} + {4^4} + ...... + {4^{90}}\\
 = \left( {4 + {4^2}} \right) + \left( {{4^3} + {4^4}} \right) + ...... + \left( {{4^{89}} + {4^{90}}} \right)\\
 = \left( {4 + {4^2}} \right) + {4^2}.\left( {4 + {4^2}} \right) + ...... + {4^{88}}.\left( {4 + {4^2}} \right)\\
 = 20 + {4^2}.20 + ..... + {4^{88}}.20\\
 = 20.\left( {1 + {4^2} + ..... + {4^{88}}} \right)\,\, \vdots \,\,20\\
A = 4 + {4^2} + {4^3} + {4^4} + ...... + {4^{90}}\\
 = \left( {4 + {4^2} + {4^3}} \right) + \left( {{4^4} + {4^5} + {4^6}} \right) + ..... + \left( {{4^{88}} + {4^{89}} + {4^{90}}} \right)\\
 = 4.\left( {1 + 4 + {4^2}} \right) + {4^4}.\left( {1 + 4 + {4^2}} \right) + ..... + {4^{88}}.\left( {1 + 4 + {4^2}} \right)\\
 = 4.21 + {4^4}.21 + ..... + {4^{88}}.21\\
 = 21.\left( {4 + {4^4} + .... + {4^{88}}} \right)\,\, \vdots \,\,21\\
2,\\
a,\\
A = 1 + 2 + {2^2} + ..... + {2^{2020}}\\
 \Leftrightarrow 2A = 2.\left( {1 + 2 + {2^2} + ..... + {2^{2020}}} \right)\\
 \Leftrightarrow 2A = 2.1 + 2.2 + {2.2^2} + ..... + {2.2^{2020}}\\
 \Leftrightarrow 2A = 2 + {2^2} + {2^3} + ..... + {2^{2021}}\\
 \Leftrightarrow 2A - A = \left( {2 + {2^2} + {2^3} + ..... + {2^{2021}}} \right) - \left( {1 + 2 + {2^2} + ..... + {2^{2020}}} \right)\\
 \Leftrightarrow A = {2^{2021}} - 1\\
 \Rightarrow A < B\\
b,\\
{5^{35}} < {5^{39}} = {5^{3.13}} = {\left( {{5^3}} \right)^{13}} = {125^{13}} < {128^{13}} = {\left( {{2^7}} \right)^{13}} = {2^{7.13}} = {2^{91}}\\
 \Rightarrow {5^{35}} < {2^{91}}
\end{array}\)