<p>If the value of \(\lim_{n \to \infty} \sum_{k=0}^{n} \dfrac{{}^n C_k}{n^k(k+3)}\) equals \(L\). Then \([L]\) is equal to:</p><p>[<strong>Note:</strong> Where \([k]\) denotes greatest integer function less than or equal to \(k\).]</p>
Step-by-Step Solution
Key Concept: Recognize that the sum ∑(nCk/n^k) is the binomial expansion of (1 + 1/n)^n, and split the summation to separate the (k+3) term, then apply the limit as n→∞ where (1 + 1/n)^n → e.
<p><strong>Step 1:</strong> Rewrite the sum using binomial theorem. Note that ∑(nCk/n^k) = (1 + 1/n)^n.</p><p><strong>Step 2:</strong> The given sum is ∑_{k=0}^{n} (nCk)/(n^k(k+3)). We need to handle the (k+3) factor. Split as: ∑_{k=0}^{n} (nCk/n^k) · 1/(k+3).</p><p><strong>Step 3:</strong> Recognize that as n→∞, the main contribution comes from the (1+1/n)^n → e expansion. For the (k+3) term: ∑_{k=0}^{n} (nCk/n^k) · 1/(k+3) = ∑_{k=0}^{n} (nCk/n^k) · (1/3 - 1/k+3) type decomposition, but more directly:</p><p><strong>Step 4:</strong> Using the integral representation: lim_{n→∞} ∑(nCk/n^k) · 1/(k+3) corresponds to ∫₀¹ e^x · x² dx (after proper transformation) = e·(1·2 - 2·1 + 2) = e(2) - 2e + 2e = 2e - 2.</p><p><strong>Step 5:</strong> Calculate: L = 2e - 2 ≈ 2(2.718) - 2 = 5.436 - 2 = 3.436</p><p>∴ [L] = <strong>3</strong></p>
Correct Answer: B