Let a_0, a_1, a_2, ldots, a_n in mathbb{R} be | vedclass

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(B) Let the sum be $S = \sum_{k=0}^n a_k C_k$. Since $a_k$ is in an arithmetic progression,$a_k = a_0 + kd$,where $d$ is the common difference. Thus,$

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$(a_0+a_n) \cdot 2^{n-1}$

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(B) Let the sum be $S = \sum_{k=0}^n a_k C_k$. Since $a_k$ is in an arithmetic progression,$a_k = a_0 + kd$,where $d$ is the common difference. Thus,$S = \sum_{k=0}^n (a_0 + kd) C_k = a_0 \sum_{k=0}^n C_k + d \sum_{k=0}^n k C_k$. We know that $\sum_{k=0}^n C_k = 2^n$ and $\sum_{k=0}^n k C_k = n 2^{n-1}$. Substituting these values,$S = a_0 2^n + d n 2^{n-1} = 2^{n-1} (2a_0 + nd)$. Since $a_n = a_0 + nd$,we have $2a_0 + nd = a_0 + (a_0 + nd) = a_0 + a_n$. Therefore,$S = (a_0 + a_n) 2^{n-1}$.

Let $a_0, a_1, a_2, \ldots, a_n \in \mathbb{R}$ be in an arithmetic progression and let $C_0, C_1, C_2, \ldots, C_n$ be the binomial coefficients. Then $\sum_{k=0}^n a_k \cdot C_k =$

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