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 $\beta_n$, by Vieta's formulas: $$\alpha_n + \beta_n = -(n+1)$$ $$\alpha_n \cdot \beta_n = n^2$$ Step 2: Expand $(\alpha_n+1)(\beta_n+1)$ using the roots' properties. We need to find the denominator of our sum. Expanding the product: $$(\alpha_n+1)(\beta_n+1) = \alpha_n\beta_n + \alpha_n + \beta_n + 1$$ Substituting the values from Step 1: $$(\alpha_n+1)(\beta_n+1) = n^2 + (-(n+1)) + 1 = n^2 - n - 1 + 1 = n(n-1)$$ Step 3: Express the general term using partial fractions. The general term of our sum is: $$\frac{1}{(\alpha_n+1)(\beta_n+1)} = \frac{1}{n(n-1)}$$ Using partial fraction decomposition: $$\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$$ Step 4: Evaluate the telescoping series. The sum becomes: $$\sum_{n=2}^{2021} \left(\frac{1}{n-1} - \frac{1}{n}\right)$$ Writing out the terms: $$\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{2020} - \frac{1}{2021}\right)$$ This is a telescoping series where consecutive terms cancel: $$= \frac{1}{1} - \frac{1}{2021} = 1 - \frac{1}{2021} = \frac{2020}{2021}$$ Step 5: Identify $a$ and $b$, then calculate $b-a$. The sum in lowest terms is $\frac{2020}{2021}$. We can verify that $\gcd(2020, 2021) = 1$ since $2021 = 2020 + 1$. Therefore: - $a = 2020$ - $b = 2021$ Thus: $$b - a = 2021 - 2020 = 1$$ The answer is **Option 1: 1**