<p>Let the three consecutive binomial coefficients be \({}^nC_r\), \({}^nC_{r+1}\) and \({}^nC_{r+2}\). Given \({}^nC_r : {}^nC_{r+1} : {}^nC_{r+2} = 2 : 15 : 70\). Find the average of \({}^{16}C_1 + {}^{16}C_2 + {}^{16}C_3\) divided by 3.</p>
Step-by-Step Solution
Key Concept: Use the ratio of consecutive binomial coefficients to find n and r, then apply the relationship between binomial coefficients to compute the required sum.
<p><strong>Step 1: Set up ratio equations from given information.</strong></p><p>Given: ${}^nC_r : {}^nC_{r+1} : {}^nC_{r+2} = 2 : 15 : 70$</p><p>This means: $\frac{{}^nC_r}{{}^nC_{r+1}} = \frac{2}{15}$ and $\frac{{}^nC_{r+1}}{{}^nC_{r+2}} = \frac{15}{70} = \frac{3}{14}$</p><p><strong>Step 2: Use the first ratio.</strong></p><p>$\frac{{}^nC_r}{{}^nC_{r+1}} = \frac{r+1}{n-r} = \frac{2}{15}$</p><p>This gives: $15(r+1) = 2(n-r)$</p><p>$15r + 15 = 2n - 2r$</p><p>$17r + 15 = 2n$ ... (Equation 1)</p><p><strong>Step 3: Use the second ratio.</strong></p><p>$\frac{{}^nC_{r+1}}{{}^nC_{r+2}} = \frac{r+2}{n-r-1} = \frac{3}{14}$</p><p>This gives: $14(r+2) = 3(n-r-1)$</p><p>$14r + 28 = 3n - 3r - 3$</p><p>$17r + 31 = 3n$ ... (Equation 2)</p><p><strong>Step 4: Solve the system of equations.</strong></p><p>From Equation 1: $2n = 17r + 15$, so $n = \frac{17r + 15}{2}$</p><p>Substituting into Equation 2: $17r + 31 = 3 \cdot \frac{17r + 15}{2}$</p><p>$2(17r + 31) = 3(17r + 15)$</p><p>$34r + 62 = 51r + 45$</p><p>$-17r = -17$</p><p>$r = 1$</p><p><strong>Step 5: Find n.</strong></p><p>From Equation 1: $2n = 17(1) + 15 = 32$, so $n = 16$</p><p><strong>Step 6: Verify the solution.</strong></p><p>${}^{16}C_1 = 16$, ${}^{16}C_2 = 120$, ${}^{16}C_3 = 560$</p><p>Ratio: $16 : 120 : 560 = 2 : 15 : 70$ ✓</p><p><strong>Step 7: Calculate the required answer.</strong></p><p>${}^{16}C_1 + {}^{16}C_2 + {}^{16}C_3 = 16 + 120 + 560 = 696$</p><p>Average divided by 3: $\frac{696}{3} = 232$</p><p><strong>∴ Answer: 232</strong></p>
Correct Answer: 232