Calculam mai intai numaratorul.
2070-2069+2068-2067+....+4-3+2-1=
=1+1+1+...+1 (de n ori)
Ca sa vedem de cate ore e 1, facem astfel:
2-1 -> n=1 -(2:2)-
4-3 -> n=2 -(4:2)-
6-5 -> n=3 -(6:2)-
8-7 -> n=4 -(8:2)-
.........
2070-2069 -> n = 1035 -(2070:2)-
=> 2070-2069+2068-2067+....+4-3+2-1= 1+1+1+...+1 (de 1035 ori) =
=1035×1 = 1035
O fractia e echiunitara cand numaratorul e egal cu numitorul, acum egalam:
[tex]1+2+3+...+n = 1035 \Rightarrow \frac{n(n+1)}{2} = 1035 \Rightarrow n(n+1) = 2070 \Rightarrow \\ \Rightarrow n^{2}+n-2070 = 0
\\ \Delta = 1+8280 = 8281 = 91^{2} \\ \\ n_{1,2} = \frac{-1\pm91}{2} \Rightarrow \left \{ {{n_{1}=-46(F)}\atop {n_{2}=45}} \right. \\ \\ \Rightarrow n = 45 [/tex]
