Question

va rog ajutor.....dau coronita

Answer 1

$\frac{1}{1+2+3+...+n} = \frac{1}{ \frac{n(n+1)}{2} } = \frac{2}{n(n+1)} = 2*(\frac{1}{n}- \frac{1}{n+1})$

$\frac{1}{1}+ \frac{1}{1+2}+ \frac{1}{1+2+3}+...+\frac{1}{1+2+3+...2016} = \\ =1+\frac{2}{2*3}+ \frac{2}{3*4}+...+\frac{2}{2016*2017} = \\ \\ =1+2*( \frac{1}{2}- \frac{1}{3})+2*( \frac{1}{3}-\frac{1}{4})+...+ 2*(\frac{1}{2016}-\frac{1}{2017}) = \\ \\ =1+2( \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017})=$

De aici se reduc toti termenii din paranteza inafara de primul si ultimul, si avem:

$=1+2*( \frac{1}{2}- \frac{1}{2017})=1+1- \frac{2}{2017}=2- \frac{2}{2017}= \frac{4034-2}{2017} = \frac{4032}{2017}$