<p>If the coefficient of the middle term in the expansion of \((1+x)^{2n+2}\) is \(\alpha\) and the coefficients of middle terms in the expansion of \((1+x)^{2n+1}\) are \(\beta\) and \(\gamma\), then relate \(\alpha\), \(\beta\), and \(\gamma\).</p>
<p>\(\alpha + \beta = \gamma\)</p>
<p>\(\beta + \gamma = \alpha\)</p>
<p>\(\alpha + \gamma = \beta\)</p>
<p>\(\alpha = \beta \gamma\)</p>
Step-by-Step Solution
Key Concept: In (1+x)^(2n+2), there is ONE middle term at position (n+1) with coefficient C(2n+2, n+1). In (1+x)^(2n+1), there are TWO middle terms at positions n and (n+1) with coefficients C(2n+1, n) and C(2n+1, n+1). The relationship emerges from Pascal's identity: C(2n+2, n+1) = C(2n+1, n) + C(2n+1, n+1).
<p><strong>Step 1:</strong> Identify middle terms in (1+x)^(2n+2) (even power).</p><p>Total terms = 2n+3. Middle term is at position (n+2), so the middle coefficient is:</p><p>α = C(2n+2, n+1)</p><p><strong>Step 2:</strong> Identify middle terms in (1+x)^(2n+1) (odd power).</p><p>Total terms = 2n+2. Two middle terms are at positions (n+1) and (n+2), so:</p><p>β = C(2n+1, n) and γ = C(2n+1, n+1)</p><p><strong>Step 3:</strong> Apply Pascal's Identity: C(n,r) + C(n,r-1) = C(n+1,r)</p><p>C(2n+1, n) + C(2n+1, n+1) = C(2n+2, n+1)</p><p><strong>Step 4:</strong> Therefore:</p><p><strong>α = β + γ</strong></p><p>∴ Answer: B</p>
Correct Answer: B