Question

Cum se rezolvă : 1*2+2*3+3*4+...+2013*2014 supra 2*4+4*6+6*8+...+4026*4028 = 1 supra x la puterea a 2 -a   ??? 

Answer 1

1*2+2*3+...+2013*2014=t
$\frac{t}{4t}= \frac{1}{4}$
=>$x^{2}$=4
=>x=2

Answer 2

$[ \frac{1*2+2*3+....+2013*2014 }{2*4+4*6+...+4026*4028} = \frac{1*2+2*3+....+2013*2014}{2*(1*2+2*3+....+2013*2014)}= \frac{1}{2} = \frac{1}{x patrat}$
de unde x patrat e 2
adik este + sau -$\sqrt{2}$