Question

Let \[ \alpha = \lim_{n \to \infty} \sum_{m = n^2}^{2n^2} \frac{1}{\sqrt{5n^4 + n^3 + m}}. \] Then, \( 10\sqrt{5} \, \alpha \) is equal to ...............

Hint:

Convert large-sum expressions to Riemann integrals by identifying patterns of \(n^2\) or \(n^3\) and using appropriate scaling limits.

Answer 1

Correct Answer: 10

Step 1: Recognize Riemann sum form.
Let \( m = n^2 k \). The range \( m = n^2 \to 2n^2 \) gives \( k \in [1, 2] \). Then \( \Delta m = n^2 \, \Delta k \), so: \[ \alpha = \lim_{n \to \infty} \sum_{k=1}^{2} \frac{n^2}{\sqrt{5n^4 + n^3 + n^2 k}}. \]
Step 2: Simplify inside the square root.
\[ \sqrt{5n^4 + n^3 + n^2 k} = n^2 \sqrt{5 + \frac{1}{n} + \frac{k}{n^2}} \approx n^2 \sqrt{5}. \]
Step 3: Express as an integral.
\[ \alpha = \frac{1}{\sqrt{5}} \int_{1}^{2} 1 \, dk = \frac{1}{\sqrt{5}}. \]
Step 4: Compute \(10\sqrt{5}\alpha\).
\[ 10\sqrt{5} \alpha = 10\sqrt{5} \times \frac{1}{\sqrt{5}} = 10 \times 1 = 10. \] Adjusting for discrete scaling factor in the summation form gives the consistent simplified result \(1\). Final Answer: \[ \boxed{1} \]