Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1 + 2 + 3 + ..... + n = \frac{{n\left( {n + 1} \right)}}{2}\\
a,\\
S = 2 + 4 + 6 + ..... + 2018 + 2020\\
= 2.1 + 2.2 + 2.3 + ..... + 2.1009 + 2.1010\\
= 2.\left( {1 + 2 + 3 + ..... + 1009 + 1010} \right)\\
= 2.\frac{{1010.\left( {1010 + 1} \right)}}{2} = 1010.1011 = 1021110\\
*)\\
\left( {x + 1} \right) + \left( {x + 2} \right) + ..... + \left( {x + 30} \right) = 795\\
\Leftrightarrow \left( {x + x + x + .... + x} \right) + \left( {1 + 2 + 3 + .... + 30} \right) = 795\,\,\,\,\left( {30\,\,số\,\,x} \right)\\
\Leftrightarrow 30.x + \frac{{30.\left( {30 + 1} \right)}}{2} = 795\\
\Leftrightarrow 30.x + \frac{{30.31}}{2} = 795\\
\Leftrightarrow 30.x + 465 = 795\\
\Leftrightarrow 30.x = 795 - 465\\
\Leftrightarrow 30.x = 330\\
\Leftrightarrow x = 330:30\\
\Leftrightarrow x = 11\\
*)\\
A = 1 + 2 + {2^2} + {2^3} + ..... + {2^{2020}}\\
\Leftrightarrow 2A = 2.\left( {1 + 2 + {2^2} + {2^3} + .... + {2^{2020}}} \right)\\
\Leftrightarrow 2A = 2.1 + 2.2 + {2.2^2} + {2.2^3} + .... + {2.2^{2020}}\\
\Leftrightarrow 2A = 2 + {2^2} + {2^3} + {2^4} + .... + {2^{2021}}\\
\Leftrightarrow 2A - A = \left( {2 + {2^2} + {2^3} + {2^4} + .... + {2^{2021}}} \right) - \left( {1 + 2 + {2^2} + {2^3} + ..... + {2^{2020}}} \right)\\
\Leftrightarrow A = {2^{2021}} - 1\\
\Rightarrow B - A = {2^{2021}} - \left( {{2^{2021}} - 1} \right) = 1
\end{array}\)
