If \sum_{k=1}^{n} a_{k}=9 and \sum_{k=1}^{n} b_{k}=6, find the following values, \[ \sum_{k=1}^{n} 2 a_{k}, \sum_{k=1}^{n} \frac{b_{k}}{6}, \sum_{k=1}^{n}\left(a_{k}+b_{k}\right), \sum_{k=1}^{n}\left(a_{k}-b_{k}\right), \sum_{k=1}^{n}\left(b_{k}-4 a_{k}\right) \]

Step 1 ：The first question is asking for the sum of 2 times each \(a_k\) from \(k=1\) to \(n\). Since we know that the sum of \(a_k\) from \(k=1\) to \(n\) is 9, we can simply multiply this sum by 2 to get the answer.

Step 2 ：Given that \(\sum_{k=1}^{n} a_{k} = 9\), we can calculate \(\sum_{k=1}^{n} 2 a_{k}\) by multiplying the sum by 2.

Step 3 ：So, \(\sum_{k=1}^{n} 2 a_{k} = 2 * 9 = 18\).

Step 4 ：Final Answer: The value of \(\sum_{k=1}^{n} 2 a_{k}\) is \(\boxed{18}\).