Răspuns:
[tex]1.~pentru~n=3,~2^{3}>2*3+1,~adevar\\2~presupunem,~ca~pt.~n=k,~2^{k}>2k+1.\\3.~ Sa~verificam~pt.~n=k+1,~2^{k+1}>2*(k+1)+1~este~adevar.\\2^{k+1}>2*(k+1)+1,~2*2^{k}>2k+1+2,~2^{k}+2^{k}>2k+1+2.\\Deoarece~2^{k}>2k+1,~si~pt.~n\geq 3,~2^{k}>2,~atunci~adunand~parte~cu~[/tex]
[tex]parte~ultimele~doua~inegalitati~adevarate~obtinem~2^{k}+2^{k}>2k+1+2,~deci~ 2*2^{k}>2k+1+2,~sau~2^{k+1}>2(k+1)+1.[/tex]
Explicație pas cu pas:
am obtinut inegalitate devarata si pentru n=k+1, deci conform metodei inductie matematica, ⇒2ⁿ>2n+1, pentru ∀n≥3.
