<p>If some three consecutive coefficients in the binomial expansion of <math>(x+1)^n</math> in powers of <i>x</i> are in the ratio <math>2:15:70</math>, then the average of these three coefficients is</p>
Step-by-Step Solution
Key Concept: Use the ratio of consecutive binomial coefficients to set up equations relating n and r, then solve the system to find the coefficients.
<p><strong>Step 1:</strong> Let the three consecutive coefficients be <math>\binom{n}{r-1}, \binom{n}{r}, \binom{n}{r+1}</math></p><p><strong>Step 2:</strong> Given: <math>\binom{n}{r-1} : \binom{n}{r} : \binom{n}{r+1} = 2:15:70</math></p><p><strong>Step 3:</strong> From <math>\frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{2}{15}</math>: <math>\frac{r}{n-r+1} = \frac{2}{15} \Rightarrow 15r = 2n - 2r + 2 \Rightarrow 2n - 17r + 2 = 0</math> ...(i)</p><p><strong>Step 4:</strong> From <math>\frac{\binom{n}{r}}{\binom{n}{r+1}} = \frac{15}{70}</math>: <math>\frac{r+1}{n-r} = \frac{15}{70} = \frac{3}{14} \Rightarrow 14r + 14 = 3n - 3r \Rightarrow 3n - 17r - 14 = 0</math> ...(ii)</p><p><strong>Step 5:</strong> Solving (i) and (ii): From (i): <math>2n = 17r - 2</math>. Substituting in (ii): <math>3 \cdot \frac{17r-2}{2} - 17r - 14 = 0</math></p><p><math>\frac{51r - 6}{2} - 17r - 14 = 0 \Rightarrow 51r - 6 - 34r - 28 = 0 \Rightarrow 17r = 34 \Rightarrow r = 2</math></p><p><strong>Step 6:</strong> From (i): <math>2n = 17(2) - 2 = 32 \Rightarrow n = 16</math></p><p><strong>Step 7:</strong> The coefficients are <math>\binom{16}{1} = 16, \binom{16}{2} = 120, \binom{16}{3} = 560</math></p><p><strong>Step 8:</strong> Average <math>= \frac{16 + 120 + 560}{3} = \frac{696}{3} = 232</math></p><p>∴ Answer is (c) 232.</p>
Correct Answer: c