Question

cum se rez x(1+1/2+1/3+1/4+....1/2014)=2014-1/2-2/3-3/4-.......2013/2014

Answer 1

x(1+1/2+1/3+1/4+...1/2014)=2014+1/2-1+1/3-1+1/4-1+.....1/2014-1=> x(1+1/2+1/3+....1/2014)=2014-2013+1/2+1/3+1/4+....1/2014)=>  x(1+1/2+1/3+1/4+...1/2014)=1+1/2+1/3+1/4+...1/2014=>x=(1+1/2+1/3+1/4+...1/2014)/1+1/2+1/3+1/4+...1/2014=>x=1

Answer 2

[tex]x(1+ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \hdots +\frac{1}{2014})=2014-\frac{1}{2}-\frac{2}{3}-\frac{3}{4}- \hdots -\frac{2013}{2014} \\ Dar ~2014 = 1+1+1 + \hdots ~de~2014~ori~\hdots +1+1+1 \\ Rezulta: \\ x(1+ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \hdots +\frac{1}{2014})= \\ = 1+(1-\frac{1}{2})+(1-\frac{2}{3})+(1-\frac{3}{4})+ \hdots +(1-\frac{2013}{2014}) [/tex]


[tex]x(1+ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \hdots +\frac{1}{2014}) = ~1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \hdots +\frac{1}{2014} \\ \\ \displaystyle x= \frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \hdots +\frac{1}{2014}}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \hdots +\frac{1}{2014}} = \boxed{1}[/tex]