we have been asked to find the sum of the given geometric series
[tex] \sum _{n=1}^4\left(\frac{1}{2}\right)^{n+1}
[/tex]
A geometric sequence has a constant ratio "r" and is given by
[tex] r=\frac{a_{n+1}}{a_n} [/tex]
[tex] a_n=\left(\frac{1}{2}\right)^{n+1},\:a_{n+1}=\left(\frac{1}{2}\right)^{\left(n+1\right)+1} [/tex]
[tex] r=\frac{\left(\frac{1}{2}\right)^{\left(n+1\right)+1}}{\left(\frac{1}{2}\right)^{n+1}}=\frac{1}{2} [/tex]
The first term of the sequence is
[tex] a_1=\left(\frac{1}{2}\right)^{1+1}=\frac{1}{4} [/tex]
Sum of the sequence is given by the formula
[tex] S_n=a_1\frac{1-r^n}{1-r} [/tex]
Plug in the values we get
[tex] S_4=\frac{1}{4}\cdot \frac{1-\left(\frac{1}{2}\right)^4}{1-\frac{1}{2}} [/tex]
On simplification we get
[tex] S_4=\frac{15}{32} [/tex]
Hence sum[tex] =\frac{15}{32} [/tex]
