Question

1)Rezolvati ecuatia: x+2+3x+....2015x=1+2+3...+2014.

Answer 1

x+2+3x+....2015x=1+2+3...+2014
x+2-2+3x+4-4+....+2014 - 2014 + 2015x = 1+3+5+..+2013
x+3x+5x+...+2015x = 1+3+5+...+2013
x(1+3+5+...+2015)=1+3+5+...+2013

aici presupun ca mai trebuie un + 2015 in enunt.
deci...
x(1+3+5+...+2015)=1+3+5+...+2013 + 2015
de aici rezulta ca x=1

Answer 2

$\displaystyle x+2+3x+....2015x=1+2+3...+2014 \\ \\ x(1+2+3+...+2015)= \frac{2014(2014+1)}{2} \\ \\ x \cdot \frac{2015(2015+1)}{2} = \frac{2014 \times 2015}{2} \\ \\ x \cdot \frac{2015 \times 2016}{2} = \frac{4058210}{2} \\ \\ x \cdot \frac{4062240}{2} =2029105 \\ \\ 2031120x=2029105 \\ \\ x= \frac{2029105}{2031120}$