Answer
a) -15
b) 1
c) 1
d) -11
e) 16
Work Step by Step
a) \begin{equation}\sum_{k=1}^{n}3a_{k}=3\sum_{k=1}^{n}a_{k}=3\times-5=-15\end{equation}
b) \begin{equation}\sum_{k=1}^{n}\frac{b_{k}}{6}=\frac{1}{6}\sum_{k=1}^{n}b_{k}=\frac{1}{6}\times6=1\end{equation}
c) \begin{equation}\sum_{k=1}^{n}(a_{k}+b_{k})=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}=-5+6=1\end{equation}
d) \begin{equation}\sum_{k=1}^{n}(a_{k}-b_{k})=\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{n}b_{k}=-5-6=-11\end{equation}
e) \begin{equation}\sum_{k=1}^{n}(b_{k}-2a_{k})=\sum_{k=1}^{n}b_{k}-2\sum_{k=1}^{n}a_{k}=6-2(-5)=16\end{equation}
