Răspuns:
a) 2²⁰²⁰; b) 6
Explicație pas cu pas:
22. a)
[tex]S = {2}^{0} + {2}^{1} + {2}^{2} + ... + {2}^{2019}[/tex]
[tex]2S = 2 \cdot ({2}^{0} + {2}^{1} + {2}^{2} + ... + {2}^{2019})[/tex]
[tex]2S = {2}^{1} + {2}^{2} + ... + {2}^{2019} + {2}^{2020}[/tex]
[tex]2S + 1 = 1 + {2}^{1} + {2}^{2} + ... + {2}^{2019} + {2}^{2020} \\ [/tex]
[tex]2S + 1 = ({2}^{0} + {2}^{1} + {2}^{2} + ... + {2}^{2019}) + {2}^{2020} \\ [/tex]
[tex]2S + 1 = S + {2}^{2020}[/tex]
[tex]\red {\bf S + 1 = {2}^{2020}}[/tex]
22. b)
puterile nenule ale lui 2 se repetă la fiecare 4 puteri consecutive:
[tex]u( {2}^{2020} ) = u( {2}^{4\cdot505}) = u( {2}^{4}) = \red{\bf 6} \\ [/tex]