Question

Problema de clasa a 7-a. Dau coroana. ​

Answer 1

$\tfrac{1}{2} + \tfrac{1}{6} + \tfrac{1}{12} + \tfrac{1}{20} + ... + \tfrac{1}{a(a + 1)} = \tfrac{2021}{2022} \\ \\ \tfrac{1}{1 \times 2} + \tfrac{1}{2 \times 3} + \tfrac{1}{3 \times 4} + ... + \tfrac{1}{a(a + 1)} = \tfrac{2021}{2022} \\ \\ \tfrac{2}{1 \times 2} - \tfrac{1}{1 \times 2} + \tfrac{3}{2 \times 3} - \tfrac{2}{2 \times 3} + ... + \tfrac{a + 1 - a}{a(a + 1)} = \tfrac{2021}{2022} \\ \\ \tfrac{1}{1} - \tfrac{1}{2} + \tfrac{1}{2} - \tfrac{1}{3} + ... + \tfrac{a + 1}{a(a + 1)} - \tfrac{a}{a(a + 1)} = \tfrac{2021}{2022} \\ \\ \tfrac{1}{1} - \tfrac{1}{a} + \tfrac{1}{a} - \tfrac{1}{a + 1} = \tfrac{2021}{2022} \\ \\ 1 - \frac{1}{a + 1} = \frac{2021}{2022} \\ \\ \frac{a + 1}{a + 1} - \frac{1}{a + 1} = \frac{2021}{2022} \\ \\ \frac{a}{a + 1} = \frac{2021}{2022} \\ \\ 2022a = 2021a + 2021 \huge |\normalsize \: - 2021a \\ 2022a - 2021a = 2021 \\\bold{\boxed{ a = 2021}}$

Answer 2

Explicație pas cu pas:

$\dfrac{1}{ {a}^{2} + a} = \dfrac{1}{a(a + 1)} = \dfrac{1}{a} - \dfrac{1}{a + 1}$

$\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + ... + \dfrac{1}{ {a}^{2} + a} = \dfrac{2021}{2022}$

$\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \dfrac{1}{4 \cdot 5} + ... + \dfrac{1}{a(a + 1)} = \dfrac{2021}{2022}$

$\dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ... + \dfrac{1}{a - 1} - \dfrac{1}{a} + \dfrac{1}{a} - \dfrac{1}{a + 1} = \dfrac{2021}{2022}$

$\dfrac{1}{1} - \dfrac{1}{a + 1} = \dfrac{2021}{2022} \iff \dfrac{a + 1 - 1}{a + 1} = \dfrac{2021}{2022}$

$\dfrac{a}{a + 1} = \dfrac{2021}{2022} \iff 2022a = 2021a + 1$

$2022a - 2021a = 2021 \implies \bf a = 2021$