Question

punctul c ex 2 pls am nevoie

Answer 1

luam prima parte a execritiului:
$x(1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{3}+....+ \frac{1}{2014})$

o rescriem in felul urmator:

$x[1+(1- \frac{1}{2} )+(1- \frac{2}{3} )+(1- \frac{3}{4} )+...+(1- \frac{2013}{2014} )]$

daca aduma toti unu o sa avem:

$x(2014- \frac{1}{2}- \frac{2}{3}- \frac{3}{4}-...- \frac{2013}{2014})$

deci x =1