Let $a_0, a_1, a_2, \ldots a_n \in R$ be in an arithmetic progression and let $C_0, C_1, C_2, \ldots, C_n$…
Let $a_0, a_1, a_2, \ldots a_n \in R$ be in an arithmetic progression and let $C_0, C_1, C_2, \ldots, C_n$ be the binomial coefficients. Then $\sum_{k=0}^n a_k \cdot C_k=$
$\frac{1}{2}\left(a_0+a_n\right)$
$\left(a_0+a_n\right) \cdot 2^{n-1}$
$\left(a_0+a_n\right)$
0
Solution
$
\begin{aligned}
\sum_{k=0}^n a_k \cdot C_k=a_0 C_0+a_1 C_1+a_2 C_2 & +\ldots+a_n C_n \\
=a_0 C_0+\left(a_0+d\right) C_1 & +\left(a_0+2 d\right) C_2 \\
& +\ldots+\left(a_0+n d\right) C_n
\end{aligned}
$
Where $d$ is an common difference of an AP
$
\begin{aligned}
= & a_0\left(C_0+C_1+\ldots+C_n\right)+ \\
& \quad d\left(C_1+2 C_2+3 C_3+\ldots+{ }^n C_n\right) \\
= & a_0 \cdot 2^n+d\left(n \cdot 2^{n-1}\right) \\
= & 2^{n-1}\left[2 a_0+n d\right]=\left(a_0+a_n\right) 2^{n-1}
\end{aligned}
$