### Instruction: Compute \[\sum_{n=1}^{1000} \frac{1}{n^2 + n}.\].### Response: I recognize that the expression in the denominator can be factored as \(n^2 + n = n(n+1)\), so I can rewrite the summand as \[\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.\]
This suggests that I can use the technique of partial fractions, where I decompose a rational function into simpler fractions that can be added or subtracted easily.
I can rewrite the sum as \[\sum_{n=1}^{1000} \left(\frac{1}{n} - \frac{1}{n+1}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{1000} - \frac{1}{1001}\right).\]
I notice that most of the terms cancel out, leaving only the first and last terms.
This is a common pattern for telescoping series, where the partial sums collapse to a constant or simple expression.