Answer
$$248$$
Work Step by Step
$$\eqalign{
& \sum\limits_{i = 1}^5 {4 \cdot {2^i}} \cr
& = \sum\limits_{i = 1}^5 {{2^2} \cdot {2^i}} \cr
& = \sum\limits_{i = 1}^5 {{2^3} \cdot {2^{i - 1}}} \cr
& = 8\sum\limits_{i = 1}^5 {{2^{i - 1}}} \cr
& {a_n} = {\left( 2 \right)^{i - 1}},\,\,\,n = 5,\,\,\,\,{a_1} = 1 \cr
& {\text{The }}n{\text{th term is }}{a_n} = {a_1}{r^{n - 1}},\, \cr
& {a_n} = {\left( 2 \right)^{i - 1}}{\text{ with }}r = 2 \cr
& \cr
& {\text{The sum of the First }}n{\text{ terms is}} \cr
& {S_n} = \frac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}}\,\,\,\left( {{\text{Where }}r \ne 1} \right) \cr
& {\text{Then,}} \cr
& {S_5} = 8\left[ {\frac{{\left( 1 \right)\left( {1 - {{\left( 2 \right)}^5}} \right)}}{{1 - \left( 2 \right)}}} \right] \cr
& {S_5} = 248 \cr} $$
