Question:

Step 1: Apply Vieta's formulas to find the sum and product of roots. For the quadratic equation $x^2 + (n+1)x + n^2 = 0$ with roots $\alpha_n$ and $\beta_n$, Vieta's formulas give us: $$\alpha_n + \beta_n = -(n+1)$$ $$\alpha_n \beta_n = n^2$$ Step 2: Expand the product $(\alpha_n + 1)(\beta_n + 1)$. We need to find the denominator of our general term. Expanding: $$(\alpha_n + 1)(\beta_n + 1) = \alpha_n \beta_n + \alpha_n + \beta_n + 1$$ Substituting the values from Vieta's formulas: $$(\alpha_n + 1)(\beta_n + 1) = n^2 + (-(n+1)) + 1 = n^2 - n - 1 + 1 = n^2 - n$$ Factoring: $$(\alpha_n + 1)(\beta_n + 1) = n(n-1)$$ Step 3: Express the general term using partial fractions. Now we can write: $$\frac{1}{(\alpha_n + 1)(\beta_n + 1)} = \frac{1}{n(n-1)}$$ Using partial fraction decomposition: $$\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$$ Step 4: Evaluate the telescoping sum. The sum becomes: $$\sum_{n=2}^{2021} \frac{1}{n(n-1)} = \sum_{n=2}^{2021}\left(\frac{1}{n-1} - \frac{1}{n}\right)$$ Expanding the telescoping series: $$= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{2020} - \frac{1}{2021}\right)$$ Most terms cancel, leaving only: $$= 1 - \frac{1}{2021} = \frac{2021 - 1}{2021} = \frac{2020}{2021}$$ Step 5: Identify $a$ and $b$, then calculate $b - a$. The sum is expressed as $\frac{a}{b} = \frac{2020}{2021}$ where $a = 2020$ and $b = 2021$. Since $\gcd(2020, 2021) = 1$, this fraction is already in lowest terms. Therefore: $$b - a = 2021 - 2020 = 1$$ The answer is **Option 1: $1$**