Answer
(a) $ -1$
(b) $ 13$
(c) $ 12$
Work Step by Step
Given
$$ b_{1}=4, b_{2}=1, b_{3}=2, \text { and } b_{4}=-4$$
Then
\begin{align*}
(a) \sum_{i=2}^{4} b_{i}&=b_{2}+b_{3}+b_{4}\\
&=1+2+(-4)=-1
\end{align*}
and
\begin{align*}
(b) \sum_{j=1}^{2}\left(2^{b_{j}}-b_{j}\right)&=\left(2^{b_1}-4\right)+\left(2^{b_2}-1\right)\\
&=\left(2^{4}-4\right)+\left(2^{1}-1\right)\\
&=13
\end{align*}
and
\begin{align*}
(c) \sum_{k=1}^{3} k b_{k}&= b_1+2b_2+3b_3\\
&=1(4)+2(1)+3(2)\\
&=12
\end{align*}
