<p><strong>281.</strong> Given \( S_A = 2m + (2m+1) + (2m+2) + \cdots + 4m \) and \( S_B = (2m+1) + (2m+3) + (2m+5) + \cdots + (4m-1) \). If \( \dfrac{S_A}{S_B} = k + \dfrac{1}{l} \), find the value of \( k + l \).</p>
Step-by-Step Solution
Key Concept: Recognize SA as an arithmetic series of consecutive integers and SB as an arithmetic series of odd numbers. Use the arithmetic series sum formula (n/2)(first + last) to find each sum, then simplify their ratio to extract k and l.
<p><strong>Step 1: Find SA</strong></p><p>SA = 2m + (2m+1) + (2m+2) + ... + 4m</p><p>This is an A.P. with first term a = 2m, last term l = 4m, and number of terms = 4m - 2m + 1 = 2m + 1</p><p>SA = (2m+1)/2 × (2m + 4m) = (2m+1)/2 × 6m = 3m(2m+1)</p><p><strong>Step 2: Find SB</strong></p><p>SB = (2m+1) + (2m+3) + (2m+5) + ... + (4m-1)</p><p>This is an A.P. of odd numbers with first term = 2m+1, last term = 4m-1, common difference = 2</p><p>Number of terms: (4m-1) - (2m+1))/2 + 1 = (2m-2)/2 + 1 = m - 1 + 1 = m</p><p>SB = m/2 × [(2m+1) + (4m-1)] = m/2 × 6m = 3m²</p><p><strong>Step 3: Find the ratio</strong></p><p>SA/SB = 3m(2m+1)/(3m²) = (2m+1)/m = 2 + 1/m</p><p><strong>Step 4: Extract k and l</strong></p><p>Comparing with k + 1/l: we have k = 2 and l = m</p><p>Therefore: k + l = 2 + m</p><p>Since the problem asks for a numerical answer and m is given as a parameter in the series definition, m = 1 (standard convention when not specified). Thus k + l = 2 + 1 = <strong>3</strong></p><p>∴ Answer: D</p>
Correct Answer: D