Explicație pas cu pas:
21.
[tex]u( {2018}^{2021} + {2019}^{2020} + {2020}^{2021} ) = u( {8}^{2021}) + u({9}^{2020}) + u({0}^{2021} ) = \\[/tex]
[tex]= u({8}^{4\cdot505} \cdot 8) + u( {9}^{2\cdot1010} ) + u(0) = u({8}^{4} \cdot 8) + u( {9}^{2} ) + 0 \\ [/tex]
[tex]= u(6 \cdot 8) + 1 = 8 + 1 = \red{\bf 9} \\ [/tex]
restul împărțirii la 10 al numărului este 9
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]
b)
[tex]u( {2}^{2020} ) = u( {2}^{4\cdot505}) = u( {2}^{4}) = u(16) = \red{\bf 6} \\ [/tex]
23.
[tex]a = 1 + {3}^{1} + {3}^{2} + ... + {3}^{2019}[/tex]
[tex]3a = 3 \cdot (1 + {3}^{1} + {3}^{2} + ... + {3}^{2019}) \\ [/tex]
[tex]3a = {3}^{1} + {3}^{2} + ... + {3}^{2019} + {3}^{2020} \\ [/tex]
[tex]3a + 1 = 1 + {3}^{1} + {3}^{2} + ... + {3}^{2019} + {3}^{2020} \\ [/tex]
[tex]3a + 1 = (1 + {3}^{1} + {3}^{2} + ... + {3}^{2019}) + {3}^{2020} \\ [/tex]
[tex]3a + 1 = S + {3}^{2020}[/tex]
[tex]2a + 1 = {3}^{2020} = {3}^{2 \cdot 1010} = ({3}^{1010})^{2} \\ [/tex]
[tex]\red {\bf 2a + 1 = ({3}^{1010})^{2}}[/tex]
