Question

Fie $(a_n)_{n\geq 1}$ un sir de numere reale definit astfel $a_1=1010$ si $\frac{a_1+a_2+a_3+...+a_{n-1}}{a_n} =n^2 -1$, pentru n≥2.
Sa se calculeze $a_{2019}$

E din GM nr 1/2019 cls 10


Daca se poate o rezolvare amanuntita sau ideile de rezolvare. Multumesc anticipat!

Answer 1

$\dfrac{a_1+a_2+a_3+...+a_{n-1}}{a_n} = n^2-1\Big|+\dfrac{a_n}{a_n} \\ \\ \dfrac{a_1+a_2+a_3+...+a_{n-1}}{a_n}+\dfrac{a_n}{a_n} = n^2-1+1 \\ \\ \dfrac{a_1+a_2+a_3+...+a_n}{a_n} = n^2 \\\\ n = 2 \\ \\ \dfrac{a_1+a_2}{a_2} = 4 \Rightarrow a_2 = \dfrac{1010}{3} \\ \\ n = 3 \\ \\ \dfrac{a_1+a_2+a_3}{a_3} = 9 \Rightarrow a_3 = \dfrac{1010}{6}$

$n=4 \\ \\ \dfrac{a_1+a_2+a_3+a_4}{a_4} = 16 \Rightarrow a_4 = \dfrac{1010}{10} \\ \\ \text{Avem sirul 1,3,6,10,...} \\ \\ \Rightarrow a_n = \dfrac{1010}{\dfrac{n(n+1)}{2}} \Rightarrow a_n = \dfrac{2020}{n(n+1)}\\ \\\\ a_{2019} = \dfrac{2020}{2019\cdot 2020} \Rightarrow \boxed{a_{2019} = \dfrac{1}{2019}}$