Question

Va rooggggg!!! Exercițiul 2c !!!!
Dau coroana!!!!!

Answer 1

$\it 2015=\underbrace{1+1+1+\ ...\ +1}_{2015\ termeni}\\ \\ \\ Membrul\ drep\ al\ ecua\c{\it t}iei\ se\ poate\ scrie:\\ \\ 1+\left(1-\dfrac{1}{2} \right) +\left(1-\dfrac{2}{3} \right) +\left(1-\dfrac{3}{4} \right)+\ ...\ +\left(1-\dfrac{2014}{2015} \right) =\\ \\ \\ =1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\ ...\ +\dfrac{1}{2015}$

Acum, ecuația devine:

$\it x\cdot\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\ ...\ +\dfrac{1}{2015}\right)=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\ ...\ +\dfrac{1}{2015}\right) \Rightarrow\\ \\ \\ \Rightarrow x=1.$

Answer 2

Răspuns

Explicație pas cu pas:

Notam cu S suma 1+1/2+1/3+1/4+...+1/2015.

Calculam separat membrul drept:

  2015-1/2-2/3-3/4-...-2014/2015=

= 1+(1-1/2)+(1-2/3)+(1-3/4)+...+(1-2014/2015)=

=1+(2/2-1/2)+(3/3-2/3)+(4/4-3/4)+...+(2015/2015-2014/2015)=

=1+ 1/2+1/3+1/4+...+1/2015=S

=> ecuatia devine: x·S=S <=> x=S/S <=>  x=1.