Question

nu sm nicio idee.. ma gandeam la inegalitatea mediilor sau la ceva cu progresii insa nu stiu sa aplic.

Answer 1

$\displaystyle Se~poate~direct;~prin~eliminari~de~variante.~Voi~rezolva~prima~data \\ \\ "cinstit"~problema,~apoi~voi~rezolva~"necinstit". ~In~realitate~metoda \\ \\ "necinstita"~este~pasul~care~completeaza~metoda~"cinstita".\\ \\ 1.~ Metoda~"cinstita" \\\\ Avem~\frac{1}{\sqrt k}= \frac{2}{2 \sqrt k}= \frac{2}{\sqrt k + \sqrt k}< \frac{2}{\sqrt{k-1}+\sqrt k}=2( \sqrt{k}- \sqrt{k-1})~\forall~k \ge 1. \\ \\ Deci: \\ \\ S<2(\sqrt 1- \sqrt 0)+2(\sqrt 2 -\sqrt 1 )+2(\sqrt 3 - \sqrt 2)+...+2(\sqrt{n}-\sqrt{n-1})$

$\displaystyle \Leftrightarrow S_n<2 \sqrt n~\forall~n \ge 2. \\ \\ Deci~c)~functioneaza. \\ \\ ---------- \\ \\ In~mod~normal~faptul~ca~2 \sqrt{n}~functioneaza~nu~garanteaza~ca~nu \\ \\ ar~mai~putea~fi~si~alte~variante~bune. ~Deci~pentru~ca~solutia~sa~fie \\ \\ completa~ar~mai~trebui~sa~demonstram~ca~celelalte~variante~nu \\ \\ functioneaza,~ceea~ce~este~de~fapt~asa-numita~metoda~"necinstita" \\ \\ (eliminare~de~variante).$

$\displaystyle 2.~Metoda~"necinstita" \\ \\ Cine~stie~din~start~ca~suma~respectiva~este~nemarginita,~poate~elimina \\ \\ variantele~a,b,d.~(E~cunoscut:~S_n> 1+ \frac{1}{2}+ \frac{1}{3}+...+ \frac{1}{n} \to \infty). \\ \\Variante~e)~esueaza~in~mod~evident~la~testul~n=2. \\ \\ Asadar~si~varianta~f)~esueaza~(caci~\frac{n}{3}< \frac{n}{2}). \\ \\ Ramane~c).$