Answer
$115$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To evaluate the given expression, $
\displaystyle\sum_{i=1}^{5} (8i-1)
,$ use the properties of summation.
$\bf{\text{Solution Details:}}$
Using a property of the summation which is given by $\displaystyle\sum_{i=1}^n (a_i-b_i)=\displaystyle\sum_{i=1}^n a_i-\displaystyle\sum_{i=1}^nb_i,$ with $n=
5
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\displaystyle\sum_{i=1}^{5} 8i-\displaystyle\sum_{i=1}^{5}1
.\end{array}
Using a property of the summation which is given by $\displaystyle\sum_{i=1}^n ca_i=c\displaystyle\sum_{i=1}^n a_i,$ with $c=
8
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
8\displaystyle\sum_{i=1}^{5} i-\displaystyle\sum_{i=1}^{5}1
.\end{array}
Using a property of the summation which is given by $\displaystyle\sum_{i=1}^n i=\dfrac{n(n+1)}{2},$ with $n=
5
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
8\left( \dfrac{5(5+1)}{2}\right)-\displaystyle\sum_{i=1}^{5}1
\\\\=
8\left( \dfrac{5(6)}{2}\right)-\displaystyle\sum_{i=1}^{5}1
\\\\=
8\left( 15\right)-\displaystyle\sum_{i=1}^{5}1
\\\\=
120-\displaystyle\sum_{i=1}^{5}1
.\end{array}
Using a property of the summation which is given by $\displaystyle\sum_{i=1}^n c=nc,$ with $n=
5
,$ and $c=
1
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
120-5(1)
\\\\=
120-5
\\\\=
115
.\end{array}
