Question

Rezolvati ecuatia:$\frac{x}{2} + \frac{x+1}{5} + \frac{x+2}{8} +....+ \frac{x+33}{101} = \frac{1}{x+1} + \frac{2}{x+4} + \frac{3}{x+7} +...+ \frac{34}{x+100}$

Answer 1

$Deoarece~ x \in N^* \Rightarrow x \geq 1.~( Fara~aceasta~informatie~problema~nu \\ \\ poate~fi~rezolvata!~Am~spus-o~si~o~repet:~sunt~diverse~ecuatii~ \\ \\ specifice~unor~multimi,~adica~nu~pot~fi~rezolvate~decat~in~ \\ \\ multimile~respective!~Exista~ecuatii~care~se~rezolva~utilizand~ \\ \\ inegalitati,~ca~problema~de~fata,~iar~inegalitatile~clasice~nu~sunt~ \\ \\ valabile~pentru~orice~numere...~Jur~ca~daca~mai~vad~ca~o~multime \\ \\nu~este~precizata,~nu~te~mai~ajut~cu~problemele!)$

$REVENIND:~Voi~nota~E(x)= \frac{x}{2}+ \frac{x+1}{5}+ \frac{x+2}{8}+...+ \frac{x+33}{101}~si \\ \\ F(x)= \frac{1}{x+1}+ \frac{2}{x+4}+ \frac{3}{x+7}+...+ \frac{34}{x+100}. \\ \\ Si~cum~x \geq 1~rezulta~E(x) \geq E(1)~(numarator~mai~mare-fractie~ \\ \\ mai~mare)~si~F(x) \leq F(1)~(numitor~mai~mare-fractie~mai~mica) . \\ \\Se~constata~ca~E(1)=F(1):~termenii~lui~E(x)~sunt~de~forma ~ \frac{x+k}{3k+2}, \\ \\ iar~termenii~lui~F(x)~sunt~de~forma~ \frac{k+1}{x+1+3k} ~(k \in \{0;1;2...33\}) .$

$Pentru~x=1~avem~ \frac{x+k}{3k+2} = \frac{1+k}{3k+2},~iar~ \frac{k+1}{x+3k+1}= \frac{1+k}{3k+2},~deci~x=1 \Rightarrow \\ \\ \Rightarrow E(x)~si~F(x)~au~termenii~egali~( \frac{x}{2}= \frac{1}{x+1}~;~ \frac{x+1}{5}= \frac{2}{x+4},~s.a.m.d.). \\ \\ De~aceea~E(1)=F(1) \Rightarrow E(x) \geq E(1) =F(1) \geq F(x). \\ \\ Rezulta~ca~egalitatea~E(x)=F(x)~are~loc~\Leftrightarrow \boxed{x=1}$