Question

a= 1/1×2 + 1/2×3 + 1/3×4 +
.....+ 1/ 2022×2023 .Stabiliți daca a€ (6/7; 1)​

Answer 1

Explicație pas cu pas:

$a = \dfrac{1}{1\cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + ... + \dfrac{1}{2021 \cdot 2022} + \dfrac{1}{2022 \cdot 2023} =$

$= \dfrac{2 - 1}{1\cdot 2} + \dfrac{3 - 2}{2 \cdot 3} + \dfrac{4 - 3}{3 \cdot 4} + ... + \dfrac{2022 - 2021}{2021 \cdot 2022} + \dfrac{2023 - 2022}{2022 \cdot 2023}$

$= \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{2021} - \dfrac{1}{2022} + \dfrac{1}{2022} - \dfrac{1}{2023}$

$= \dfrac{1}{1} - \dfrac{1}{2023} = \dfrac{2023 - 1}{2023} = \bf \dfrac{2022}{2023}$

$\dfrac{6}{7} = \dfrac{1734}{2023} < \dfrac{2022}{2023} < 1$

$\implies a \in \Big(\dfrac{6}{7};1\Big)$