Demonstrație:
[tex]a = 1 + 3 + {3}^{2} + {3}^{3} + ... + {3}^{2019} \\ 3a = 3 + {3}^{2} + {3}^{3} + {3}^{4} + ... + {3}^{2020} \\ - - - - - - - - - - - - - - \\ 3a - a = (3 + {3}^{2} + {3}^{3} + ... + {3}^{2020} ) - (1 + 3 + {3}^{2} + ... {3}^{2019} ) \\ 2a = {3}^{2020} - 1 \\ 2a + 1 = {3}^{2020} - 1 + 1 = {3}^{2020} \\ 2a + 1 = ( {3}^{1010} )^{2} ⇔2a + 1 \: este \: p.p.[/tex]
