Question

Sa se demonstreze ca daca n>=5 este nr natural,atunci 2 la putrea n >n la putera a 2a +n+1

Answer 1

$\displaystyle Vom~demonstra~aceasta~afirmatie~prin~inductie. \\ \\ Etapa~de~demonstratie:~Pentru~n=5~propozitia~este~adevarata, \\ \\ fiind~echivalenta~cu~2^5 \geq 5^2+5+1 \Leftrightarrow 32 \geq 31,~adevarat! \\ \\ Etapa~de~demonstratie:~Presupunem~ca~propozitia~este \\ \\ adevarata~pentru~un~numar~natural~k \geq 5,~si~demonstram~ca \\ \\ este~adevarata~si~pentru~k+1. \\ \\ Mai~exact,~presupunem~ca~2^k \geq k^2+k+1,~si~demonstram~ca \\ \\ 2^{k+1} \geq (k+1)^2+(k+1)+1.$

$\displaystyle Din~2^k \geq k^2+k+1,~rezulta~(prin~inmultire~cu~2)~ca \\ \\ 2^{k+1} \geq 2k^2+2k+2.........(1) \\ \\ In~continuare~ne~propunem~sa~demonstram~ca~ \\ \\ 2k^2+2k+2 \geq (k+1)^2+(k+1)+1. \\ \\ Ultima~relatie~este~echivalenta~succesiv~cu: \\ \\ 2k^2+2k+2 \geq k^2+2k+1+k+1+1 \Leftrightarrow k^2-k-2 \geq 0 \Leftrightarrow \\ \\ \Leftrightarrow (k+1)(k-2) \geq 0,~adevarat,~caci~k \geq 5........(2). \\ \\ Din~(1)~si~(2)~rezulta~2^{k+1} \geq (k+1)^2+(k+1)+1,~q.e.d.$