<p>If the coefficients of \(r^{\text{th}}\) and \((r+1)^{\text{th}}\) terms in the expansion of \((3+7x)^{29}\) are equal, then \(r\) equals</p>
Step-by-Step Solution
Key Concept: In the binomial expansion of (a+bx)^n, the coefficient of the (r+1)th term is C(n,r)·a^(n-r)·b^r. Set coefficients of consecutive terms equal and use the relationship between consecutive binomial coefficients to find r.
<p><strong>Step 1:</strong> The general term in (3+7x)^29 is T_(k+1) = C(29,k)·3^(29-k)·(7x)^k</p><p><strong>Step 2:</strong> Coefficient of r-th term: C(29,r-1)·3^(30-r)·7^(r-1)</p><p><strong>Step 3:</strong> Coefficient of (r+1)-th term: C(29,r)·3^(29-r)·7^r</p><p><strong>Step 4:</strong> Set them equal: C(29,r-1)·3^(30-r)·7^(r-1) = C(29,r)·3^(29-r)·7^r</p><p><strong>Step 5:</strong> Simplify: C(29,r-1)·3/7 = C(29,r)</p><p><strong>Step 6:</strong> Using C(n,r-1)/C(n,r) = r/(n-r+1): [r/(30-r)]·(3/7) = 1</p><p><strong>Step 7:</strong> 3r = 7(30-r) → 3r = 210 - 7r → 10r = 210 → r = 21</p><p>∴ Answer: B (r = 21)</p>
Correct Answer: B