<p>If the coefficients of \(r^{\text{th}}\), \((r+1)^{\text{th}}\) and \((r+2)^{\text{th}}\) terms in the binomial expansion of \((1+y)^m\) are in AP, then \(m\) and \(r\) satisfy the equation</p>
<p>\(m^2 - m(4r-1) + 4r^2 - 2 = 0\)</p>
<p>\(m^2 - m(4r+1) + 4r^2 + 2 = 0\)</p>
<p>\(m^2 - m(4r+1) + 4r^2 - 2 = 0\)</p>
<p>\(m^2 - m(4r-1) + 4r^2 + 2 = 0\)</p>
Step-by-Step Solution
Key Concept: The binomial coefficients C(m,r-1), C(m,r), C(m,r+1) are in AP when 2·C(m,r) = C(m,r-1) + C(m,r+1). Use the relationship between consecutive binomial coefficients to convert this into a quadratic equation in m and r.
<p><strong>Step 1:</strong> Identify coefficients. The rth, (r+1)th, and (r+2)th terms in (1+y)^m have coefficients C(m,r-1), C(m,r), and C(m,r+1) respectively.</p><p><strong>Step 2:</strong> Apply AP condition. If these are in AP: 2·C(m,r) = C(m,r-1) + C(m,r+1)</p><p><strong>Step 3:</strong> Use the relation C(m,k-1) = C(m,k)·k/(m-k+1). Express C(m,r-1) = C(m,r)·r/(m-r+1) and C(m,r+1) = C(m,r)·(m-r)/(r+1)</p><p><strong>Step 4:</strong> Substitute into the AP condition: 2·C(m,r) = C(m,r)·r/(m-r+1) + C(m,r)·(m-r)/(r+1)</p><p><strong>Step 5:</strong> Divide by C(m,r): 2 = r/(m-r+1) + (m-r)/(r+1)</p><p><strong>Step 6:</strong> Multiply through by (m-r+1)(r+1): 2(m-r+1)(r+1) = r(r+1) + (m-r)(m-r+1)</p><p><strong>Step 7:</strong> Expand: 2(mr + m - r² - r + r + 1) = r² + r + m² - rm - m + rm - r² + m - r</p><p><strong>Step 8:</strong> Simplify: 2mr + 2m + 2 - 2r² = m² - r</p><p><strong>Step 9:</strong> Rearrange: m² - 2mr - 2m - 2r² + r + 2 = 0 or equivalently m² - 2m(r+1) - 2r² + r + 2 = 0</p><p>∴ Answer: <strong>m² - 2mr - 2m - 2r² + r + 2 = 0</strong> (or equivalent form depending on option C)</p>
Correct Answer: C