<p>The value of \(\displaystyle\sum_{r=1}^{15} r^2 \left(\frac{{}^{15}C_r}{{}^{15}C_{r-1}}\right)\) is equal to</p>
Step-by-Step Solution
Key Concept: Simplify the ratio of binomial coefficients to get $\frac{15C_r}{15C_{r-1}} = \frac{16-r}{r}$, then use this to transform the sum into a form involving simpler terms that telescope or factor nicely.
<p><strong>Step 1:</strong> Simplify the ratio of consecutive binomial coefficients.</p><p>$$\frac{{}^{15}C_r}{{}^{15}C_{r-1}} = \frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(16-r)!}} = \frac{(r-1)!(16-r)!}{r!(15-r)!} = \frac{16-r}{r}$$</p><p><strong>Step 2:</strong> Substitute into the sum.</p><p>$$\sum_{r=1}^{15} r^2 \cdot \frac{16-r}{r} = \sum_{r=1}^{15} r(16-r) = \sum_{r=1}^{15} (16r - r^2)$$</p><p><strong>Step 3:</strong> Split and use standard formulas.</p><p>$$= 16\sum_{r=1}^{15} r - \sum_{r=1}^{15} r^2 = 16 \cdot \frac{15 \cdot 16}{2} - \frac{15 \cdot 16 \cdot 31}{6}$$</p><p>$$= 16 \cdot 120 - 1240 = 1920 - 1240 = 680$$</p><p>∴ Answer: <strong>680</strong></p>
Correct Answer: D