<p>Given \( \displaystyle\sum_{i=1}^{20} \left(\dfrac{{}^{20}C_{i-1}}{{}^{20}C_i + {}^{20}C_{i-1}}\right)^3 = \dfrac{k}{21} \), find the value of \( k \).</p>
Step-by-Step Solution
Key Concept: Use the symmetry property ${}^{20}C_r = {}^{20}C_{20-r}$ to pair terms, and recognize that $\frac{{}^{20}C_{i-1}}{{}^{20}C_i + {}^{20}C_{i-1}} + \frac{{}^{20}C_i}{{}^{20}C_{i+1} + {}^{20}C_i} = 1$ when indices are properly matched through the complement relationship.
<p><strong>Step 1:</strong> Let $a_i = \frac{{}^{20}C_{i-1}}{{}^{20}C_i + {}^{20}C_{i-1}}$. Using ${}^{20}C_i = \frac{20!}{i!(20-i)!}$, we get:</p><p>$$a_i = \frac{{}^{20}C_{i-1}}{{}^{20}C_i + {}^{20}C_{i-1}} = \frac{i}{21}$$</p><p>This is because ${}^{20}C_i + {}^{20}C_{i-1} = {}^{21}C_i$ and $\frac{{}^{20}C_{i-1}}{{}^{21}C_i} \cdot {}^{21}C_i = {}^{20}C_{i-1}$, leading to $a_i = \frac{i}{21}$.</p><p><strong>Step 2:</strong> Therefore:</p><p>$$\sum_{i=1}^{20} a_i^3 = \sum_{i=1}^{20} \left(\frac{i}{21}\right)^3 = \frac{1}{21^3}\sum_{i=1}^{20} i^3$$</p><p><strong>Step 3:</strong> Using the formula $\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2$:</p><p>$$\sum_{i=1}^{20} i^3 = \left[\frac{20 \cdot 21}{2}\right]^2 = (210)^2 = 44100$$</p><p><strong>Step 4:</strong> Thus:</p><p>$$\sum_{i=1}^{20} a_i^3 = \frac{44100}{9261} = \frac{44100}{9261} = \frac{2100}{441} = \frac{100}{21}$$</p><p>∴ <strong>Answer:</strong> $k = \boxed{100}$</p>
Correct Answer: 100