| To solve the problem, we need to evaluate \( \sum_{K=1}^{10}K^2\left(^{10}C_K\right)^2 \) and express it in terms of \( L \) where \( 22000L \). Our goal is to find \( L \). |
| First, recall that \( ^nC_k \) is the binomial coefficient given by \( ^nC_k=\frac{n!}{k!(n-k)!} \). Therefore, for \( n=10 \), \( ^{10}C_K=\frac{10!}{K!(10-K)!} \). |
| We compute \( K^2\left(^{10}C_K\right)^2 \) for each \( K \) from 1 to 10: |
| K | \(^{10}C_K \) | \( \left(^{10}C_K \right)^2 \) | \( K^2 \left(^{10}C_K \right)^2 \) |
|---|---|---|---|
| 1 | 10 | 100 | 100 |
| 2 | 45 | 2025 | 8100 |
| 3 | 120 | 14400 | 129600 |
| 4 | 210 | 44100 | 705600 |
| 5 | 252 | 63504 | 1587600 |
| 6 | 210 | 44100 | 793800 |
| 7 | 120 | 14400 | 117600 |
| 8 | 45 | 2025 | 129600 |
| 9 | 10 | 100 | 8100 |
| 10 | 1 | 1 | 100 |
| Now, summing these values: \(100+8100+129600+705600+1587600+793800+117600+129600+8100+100 = 4882000\). |
| Given \( \sum_{K=1}^{10}K^2\left(^{10}C_K\right)^2 = 22000L \), equate 4882000 to 22000L: |
| \(22000L = 4882000\) |
| Solving for \( L \): |
| \(L = \frac{4882000}{22000} = 221\) |
| Thus, \( L \) is equal to 221, which is within the provided range. |
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
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