Answer:
5 [tex]\frac{1}{4}[/tex]
Step-by-step explanation:
The sum to n terms of a geometric sequence is
[tex]S_{n}[/tex] = [tex]\frac{a(1-r^{n}) }{1-r}[/tex]
where a is the first term and r the common ratio
The expression inside the summation is
4 [tex](\frac{1}{4}) ^{n-1}[/tex] ← the nth term of a geometric sequence
with a = 4 and r = [tex]\frac{1}{4}[/tex] , then
[tex]S_{4}[/tex] = [tex]\frac{4(1-(\frac{1}{4}) ^{3}) }{1-\frac{1}{4} }[/tex]
= [tex]\frac{4(1-\frac{1}{64}) }{\frac{3}{4} }[/tex]
= [tex]\frac{16}{3}[/tex] ( [tex]\frac{63}{64}[/tex] )
= [tex]\frac{21}{4}[/tex]
= 5 [tex]\frac{1}{4}[/tex]
