Question

afla anul descoperirii meteoritului din Hopa știind că este dat de soluția ecuației
$\frac{1}{1 \times 2 } + \frac{1}{3 \times 4} + ... + \frac{1}{(n - 1) \times n} = \frac{1919}{1920}$
n € N*​

Answer 1

$\displaystyle\\Vom~incerca~sa~scriem~pe~~\frac{1}{n(n-1)}~~ca~~\frac{A}{n}-\frac{B}{n-1},~A,B\in\mathbb{N},~n\in\mathbb{N}-\left\{0,1\right\}.\\\frac{1}{n(n-1)}=\frac{A}{n}-\frac{B}{n-1} \Longleftrightarrow \frac{A(n-1)-Bn}{n(n-1)}=\frac{1}{n(n-1)} \Longleftrightarrow\\A(n-1)-Bn=1 \Longleftrightarrow An-A-Bn=(A-B)n-A=1,~dar,~1=0\cdot n -(-1),\\\Longleftrightarrow \begin{cases} A-B=0 && A=-1 \end{cases} \Longleftrightarrow \begin{cases} A=-1 && B=1 \end{cases},$

$\displaystyle\\Deci,~\boxed{\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}}~,~acum~vom~scrie~fiecare~termen~al~sumei\\sub~forma~de~mai~sus.\\\frac{1}{1\cdot 2}=\frac{1}{1}-\frac{1}{2}.\\\frac{1}{2\cdot 3}=\frac{1}{2}-\frac{1}{3}.\\\frac{1}{3\cdot 4}=\frac{1}{3}-\frac{1}{4}.\\.\\.\\.\\\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}.\\---------(+)--\\\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n(n-1)}=1-\frac{1}{n}.$

$\displaystyle\\Deci,~avem~de~rezolvat~1-\frac{1}{n}=\frac{1919}{1920} \Longleftrightarrow \frac{n-1}{n}=\frac{1919}{1920} \Longleftrightarrow \boxed{n=1920}.\\Anul~descoperirii~meteoritului~din~Hopa~este~\boxed{1920}.$