Question

B vă rog, îmi trebuie urgent!

Answer 1

Răspuns:

n = 2021

Explicație pas cu pas:

E(x) = (-x + x²)² + 4x³ = [x(x-1)]² + 4x³ = x²(x - 1)² + 4x³ = = x²[(x - 1)² + 4x] = x²(x² - 2x + 1 + 4x) = x²(x + 1)²

=>

E(1) = 1²×2²

E(2) = 2²×3²

E(3) = 3²×4²

...

E(n) = n²(n + 1)²

$\frac{1}{ \sqrt{E(1)} } + \frac{1}{ \sqrt{E(2)} } + \frac{1}{ \sqrt{E(3)} } + ... + \frac{1}{ \sqrt{E(n)} } = \frac{2021}{2022}$

$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + ... + \frac{1}{n(n + 1)} = \frac{2021}{2022}$

$(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ... + (\frac{1}{n} - \frac{1}{n + 1}) = \frac{2021}{2022}$

$\frac{1}{1} - \frac{1}{n + 1} = \frac{2021}{2022} \\ \frac{n}{n + 1} = \frac{2021}{2022} = > n = 2021$