Binomial Theorem
Properties of Binomial Coefficients
Grade 11
Question:
<p>Let <em>T</em><sub>2−1</sub>, <em>T<sub>r</sub></em>, <em>T</em><sub><em>r</em>+1</sub> be the three successive terms of <em>(1 + x)<sup>n</sup></em>. If <sup>n</sup>C<sub>r−1</sub> : <sup>n</sup>C<sub>r−1</sub> : <sup>n</sup>C<sub>r</sub> = 1 : 7 : 42, then the first of the three given terms will be the <em>k</em>th term. Find <em>k</em>.</p>
<p>(1) 7th</p>
<p>(2) 7th</p>
<p>(3) 8th</p>
<p>(4) 6th</p>
Step-by-Step Solution
Key Concept: Use the ratio of consecutive binomial coefficients to establish equations: nCr−1/nCr = (r)/(n−r+1) and nCr/nCr+1 = (r+1)/(n−r). Solve these ratios simultaneously to find n and r, then identify which term number corresponds to Tr−1.
<p><strong>Step 1:</strong> Set up ratio equations using consecutive binomial coefficients.</p><p>Given: nCr−1 : nCr : nCr+1 = 1 : 7 : 42</p><p><strong>Step 2:</strong> From the first ratio: nCr/nCr−1 = 7</p><p>Using nCr/nCr−1 = (n−r+1)/r, we get: (n−r+1)/r = 7</p><p>∴ n − r + 1 = 7r</p><p>∴ n = 8r − 1 ... (i)</p><p><strong>Step 3:</strong> From the second ratio: nCr+1/nCr = 42/7 = 6</p><p>Using nCr+1/nCr = (n−r)/(r+1), we get: (n−r)/(r+1) = 6</p><p>∴ n − r = 6(r+1)</p><p>∴ n = 7r + 6 ... (ii)</p><p><strong>Step 4:</strong> Solve equations (i) and (ii):</p><p>8r − 1 = 7r + 6</p><p>∴ r = 7</p><p>∴ n = 8(7) − 1 = 55</p><p><strong>Step 5:</strong> Identify term positions in (1+x)^55:</p><p>• Tr−1 = T6 is the (r)th term = 7th term</p><p>• Tr = T7 is the (r+1)th term = 8th term</p><p>• Tr+1 = T8 is the (r+2)th term = 9th term</p><p>∴ The first of the three given terms (Tr−1) is the <strong>7th term</strong>, so k = 7</p>
Correct Answer: A