Answer:
The given sum expressed in sigma notation is
[tex]\sum\limits_{i=1}^{5}4^{i-4}[/tex]
Therefore [tex]\sum\limits_{i=1}^{5}4^{i-4}=\frac{1}{64}+\frac{1}{16}+\frac{1}{4}+1+4[/tex]
Step-by-step explanation:
Given series is [tex]\frac{1}{64}+\frac{1}{16}+\frac{1}{4}+1+4[/tex]
The given sum expressed in sigma notation is
[tex]\sum\limits_{i=1}^{5}4^{i-4}[/tex]
- Now verify that the sigma notation [tex]\sum\limits_{i=1}^{5}4^{i-4}[/tex] is correct or not
- Now expand the series
- [tex]\sum\limits_{i=1}^{5}4^{i-4}=4^{1-4}+4^{2-4}+4^{3-4}+4^{4-4}+4^{5-4}[/tex]
- [tex]=4^{-3}+4^{-2}+4^{-1}+4^{0}+4^{1}[/tex] ( using the properties [tex]a^{-m}=\frac{1}{a^m}[/tex] and [tex]a^0=1[/tex] )
- [tex]=\frac{1}{4^3}+\frac{1}{4^2}+\frac{1}{4^1}+1+4[/tex]
- [tex]=\frac{1}{64}+\frac{1}{16}+\frac{1}{4}+1+4[/tex]
Therefore
[tex]\sum\limits_{i=1}^{5}4^{i-4}=\frac{1}{64}+\frac{1}{16}+\frac{1}{4}+1+4[/tex]
