To solve this limit problem:

$\lim_{{n \to \infty}} \frac{n^2 + (n+1)^2 + \dots + (2n)^2}{n^3}$

we can break down the steps as follows:

Step 1: Simplify the Sum in the Numerator

The expression in the numerator is the sum of squares from $n^2$ to $(2n)^2$. We can represent this sum as: $\sum_{k=n}^{2n} k^2$

Step 2: Approximate the Sum Using the Integral

The sum $\sum_{k=n}^{2n} k^2$ can be approximated by integrating $x^2$ from $x = n$ to $x = 2n$ for large $n$. The integral approximation is: $\int_{n}^{2n} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{n}^{2n} = \frac{(2n)^3}{3} - \frac{n^3}{3} = \frac{8n^3}{3} - \frac{n^3}{3} = \frac{7n^3}{3}$

Thus, for large $n$, the sum $\sum_{k=n}^{2n} k^2$ is approximately $\frac{7n^3}{3}$.

Step 3: Substitute and Simplify

Substitute this approximation back into the original expression: $\lim_{{n \to \infty}} \frac{\sum_{k=n}^{2n} k^2}{n^3} \approx \lim_{{n \to \infty}} \frac{\frac{7n^3}{3}}{n^3} = \frac{7}{3}$

Conclusion

Therefore, the limit is: $\lim_{{n \to \infty}} \frac{n^2 + (n+1)^2 + \dots + (2n)^2}{n^3} = \frac{7}{3}$

Would you like a detailed breakdown of any specific part, or do you have any questions?

Here are some related questions for further exploration:

1. How would this approach change if we were summing $k^3$ instead of $k^2$?
2. What would the result be if the denominator were $n^2$ instead of $n^3$?
3. How can you formally justify using an integral to approximate the sum?
4. How does this solution relate to Riemann sums and integration?
5. What other methods could solve similar limit problems involving sums?

Tip: Using integrals to approximate large sums is a powerful technique for evaluating limits, especially when the sums are difficult to handle directly.