Verified answer

[tex]\displaystyle \sf A = \{a_1,a_2, a_3, ...\ , a_n\} \\\\ \text{conjunto das partes de A = P(A) } : \\\\ P(A) = \{x\ | \ x \subset A\}\to \text{(onde x s{\~a}o conjuntos)} \\\\\ Da{\'i}}: \\\\ n = 0 \to P(A) = \underbrace{\{\phi \}}}_{1 \ elemento\ =\ 2^0} \\\\ n=1 \to P(A) = \underbrace{\{ \sf \phi \ , \{a_1\}\}}_{2\ elementos\ = \ 2^1} \\\\\\ n = 2 \to P(A) =\underbrace{ \sf \{\phi, \{a_1\},\{a_2\}, \{a_1, a_2\} \} }_{\displaystyle 4 \ elementos \ = \ 2^2 } \\\\\\[/tex]

[tex]\sf n = 3 \to P(A) = \underbrace{ \sf \{\phi, \{a_1\},\{a_2\},\{a_3\} , \{a_1, a_2\}, \{a_1,a_3\}, \{a_2,a_3\}, \{a_1,a_2,a_3\} \} }_{\displaystyle 8 \ elementos\ = \ 2^3 } \\\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \\\\ n = k \to P(A) = \underbrace{\sf \{ \phi, \{a_1\},\{a_2\},\{a_3\} , \{a_1, a_2\}, \{a_1,a_3\}, \{a_2,a_3\},\ ...\ \}}_{\displaystyle 2^{k}\ elementos}[/tex]

[tex]\Large\boxed{\sf n = k \to n\left(P(A)\right) = 2 ^{k } \ , \ n\in\mathbb{N}\ }\checkmark[/tex]

onde :

[tex]\sf n\left(P(A)\right)[/tex] é o número de elementos do conjuntos das partes de A.

Com raciocínio análogo, prova-se que :

[tex]\sf n = k +1 \to P(A) = \underbrace{\{\underbrace{\sf \phi \ , \{a{_1}\}, \{a_2\}\ , .. }_{2^{k}}, \underbrace{\sf \{a_{(k+1)}\}, \{a_1,a_{(k+1)}\}\ , \{a_2,a_{(k+1)}\}\ , ..\ }_{\displaystyle 2^k } \} }_{\displaystyle 2\cdot 2^{k} = 2^{k+1}} \\\\\\\\ Portanto : \\\\\ \Large\boxed{\sf \ n = k+1 \to n(P(A)) = 2^{k+1}\ , n \in \mathbb{N}\ }\checkmark[/tex]