[tex]Pentru~n=10~avem~2^{10}\ \textgreater \ 10^3,~adevarat! \\ \\ Presupunem~ca~propozitia~este~adevarata~pentru~k \geq n,~k \in N~si~ \\ \\ demonstram~ca~este~adevarata~si~pentru~k+1. \\ \\ Avem~deci~2^k\ \textgreater \ k^3 \Rightarrow 2^{k+1} \ \textgreater \ 2k^3. \\ \\ Ne~ propunem~sa~demonstram~ca~2k^3\ \textgreater \ (k+1)^3.~Aceasta~inegalitate~ \\ \\este ~echivalenta~ca~k^3\ \textgreater \ 3k^2+3k+1.~Impartim~ultima~relatie~la~k^3,~\\ \\ obtinand~(k\ \textgreater \ 0)~: [/tex]
[tex]1\ \textgreater \ \frac{3}{k}+ \frac{3}{k^2}+ \frac{1}{k^3}~(*).~Dar~k \geq 10 \Rightarrow~ \frac{3}{k} \leq \frac{3}{10};~ \frac{3}{k^2} \leq \frac{3}{10^2};~ \frac{1}{k^3} \leq \frac{1}{10^3} . \\ \\ Prin~insumarea~ultimelor~trei~relatii,~deducem~ca~relatia~(*)~este \\ \\ adevarata,~si~deci~etapa~a~doua~de~inductie~este~verificata! [/tex]
