Let $(a_k)_{k=1}^n$ be a sequence of real numbers. We will work out all the steps of induction to prove:

$$\sum_{k=1}^n\sum_{j=1}^k \frac{a_j}{k}\binom{k}{j}=\sum_{k=1}^n \frac{a_k}{k}\binom{n}{k}$$

Write the sequence elements as "a_k" (without quotes) in your answer.

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