Tìm n $\frac{n-1}{n}$+ $\frac{n-2}{n}$+$\frac{n-3}{n}$+…+ $\frac{1}{n}$=2007

Đáp án: n=4015

Giải thích các bước giải:

$\begin{array}{l}
\frac{{n – 1}}{n} + \frac{{n – 2}}{n} + \frac{{n – 3}}{n} + … + \frac{1}{n} = 2007\\
 \Rightarrow 1 – \frac{1}{n} + 1 – \frac{2}{n} + 1 – \frac{3}{n} + … + \frac{{n – \left( {n – 1} \right)}}{n} = 2007\\
 \Rightarrow 1 – \frac{1}{n} + 1 – \frac{2}{n} + … + 1 – \frac{{n – 1}}{n} = 2007\\
 \Rightarrow \left( {1 + 1 + … + 1} \right) – \left( {\frac{1}{n} + \frac{2}{n} + … + \frac{{n – 1}}{n}} \right) = 2007\left( {n – 1\,so\,1} \right)\\
 \Rightarrow n – 1 – \frac{{1 + 2 + … + \left( {n – 1} \right)}}{n} = 2007\\
 \Rightarrow n – 1 – \frac{{\left( {n – 1 + 1} \right).\left( {n – 1} \right)}}{{2n}} = 2007\\
 \Rightarrow n – 1 – \frac{{n – 1}}{2} = 2007\\
 \Rightarrow n – 1 = 4014\\
 \Rightarrow n = 4015
\end{array}$