Question

Dau CORONITA!(rezolvarii complete+sa imi explicatii fiecare pas facut de voi)!

Answer 1

[tex]S=1+ \frac{1}{1+2} + \frac{1}{1+2+3} +.....+ \frac{1}{1+2+3+....+n} = \frac{4022}{2012} \\ \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} ) \\ [/tex]

acum inlocuim in relatia de mai sus si observam ca
pentru n=1 avem :
$2( \frac{1}{1} - \frac{1}{2} )=2* \frac{1}{2} =1$
ceea ce este adevarat
asa procedam pentru toate numerele(n=2,n=3,.....), pana la n, deci suma mea va deveni:
[tex]S=2( \frac{1}{2}- \frac{1}{2} +\frac{1}{2}- \frac{1}{3}+\frac{1}{3}-\frac{1}{4 }+....+ \frac{1}{n}- \frac{1}{n+1} ) \\ S=2(1-\frac{1}{n+1} )= \frac{4022}{2012} \\ 1-\frac{1}{n+1}=\frac{2011}{2012} \\ 2012n-2011n=2011 \\ n=2011[/tex]