Olá, Eduardo;
[tex](i)\ n=1:n^3=1^3=1,\ [\frac{n(n+1)}2]^2=[\frac{1(1+1)}2]^2=[\frac{1\cdot 2}2]^2=1[/tex]
[tex](ii)\text{ Se a proposi\c{c}\~ao\ \'e v\'alida para n} \Rightarrow \text{\'e v\'alida para n+1}[/tex]
[tex]1+8+...+n^3+(n+1)^3=[\frac{n(n+1)}2]^2+(n+1)^3=\\\\ =\frac{n^2(n^2+2n+1)}4+\frac{4n^3+12n^2+12n+4}4=\frac{n^4+2n^3+n^2+4n^3+12n^2+12n+4}4=\\\\ =\frac{n^4+4n^3+4n^2+2n^3+8n^2+8n+n^2+4n+4}4=\frac{(n^2+2n+1)(n^2+4n+4)}4=\\\\ =\frac{(n+1)^2(n+2)^2}4=[\frac{(n+1)(n+2)}2]^2[/tex]
Demonstrado.
