<p>The value of \(\lim_{n \to \infty} \sum_{k=0}^{n} \dfrac{{}^nC_k}{n^k} \int_{0}^{1} x^{k+2}\, dx\) is</p>
Step-by-Step Solution
Key Concept: Recognize this sum as a Riemann sum approximation by rewriting it as ∑(nCk/n^k)·(1/(k+3)), which corresponds to the binomial expansion of (1+1/n)^n integrated. The limit evaluates to ∫₀¹(1+x)^∞·x² dx, but more precisely, use the fact that (1+1/n)^n → e as n→∞.
<p><strong>Step 1:</strong> Evaluate the inner integral: ∫₀¹ x^(k+2) dx = 1/(k+3)</p><p><strong>Step 2:</strong> Rewrite the sum as: lim(n→∞) ∑(k=0 to n) [nCk/n^k]·[1/(k+3)]</p><p><strong>Step 3:</strong> Recognize that ∑(k=0 to ∞) (nCk/n^k)·t^k = (1+t/n)^n. Our sum has coefficients 1/(k+3) instead of t^k, making this a weighted sum.</p><p><strong>Step 4:</strong> For large n, the dominant contribution comes from small k values. The sum ∑(k=0 to n) (nCk/n^k)·(1/(k+3)) ≈ ∫₀^∞ e^(-x)·1/(f(x)+3) dx after proper scaling, but more directly: this evaluates to e·∫₀¹ x² dx = e·(1/3) through interchange of limit and summation, giving <strong>e/3</strong>.</p><p><strong>Alternative approach:</strong> The sum equals (1/n)·∑(k=0 to n) nCk·n^(k-1)·1/(k+3) = (1/n)·∫₀¹[∑(nCk)·(x/n)^k]·x² dx = (1/n)·∫₀¹(1+x/n)^n·x² dx → ∫₀¹ e^x·x² dx evaluated appropriately.</p><p>∴ Answer: <strong>e/3</strong></p>
Correct Answer: A