Question

Geometric Series Assistance

Answer 1

we have been asked to find the sum of the given geometric series

$\sum _{n=1}^4\left(\frac{1}{2}\right)^{n+1}$

A geometric sequence has a constant ratio "r" and is given by

$r=\frac{a_{n+1}}{a_n}$

$a_n=\left(\frac{1}{2}\right)^{n+1},\:a_{n+1}=\left(\frac{1}{2}\right)^{\left(n+1\right)+1}$

$r=\frac{\left(\frac{1}{2}\right)^{\left(n+1\right)+1}}{\left(\frac{1}{2}\right)^{n+1}}=\frac{1}{2}$

The first term of the sequence is

$a_1=\left(\frac{1}{2}\right)^{1+1}=\frac{1}{4}$

Sum of the sequence is given by the formula

$S_n=a_1\frac{1-r^n}{1-r}$

Plug in the values we get

$S_4=\frac{1}{4}\cdot \frac{1-\left(\frac{1}{2}\right)^4}{1-\frac{1}{2}}$

On simplification we get

$S_4=\frac{15}{32}$

Hence sum$=\frac{15}{32}$