Answer:
[tex]\sum _{n=1}^{10}8\left(\frac{1}{4}\right)^{n-1}\approx 10.67[/tex]
Step-by-step explanation:
A geometric sequence has a constant ratio [tex]r[/tex] and is defined by [tex]a_n=a_0\cdot r^{n-1}[/tex]
Check whether the ratio is constant by computing the ratios of all the adjacent terms, i.e. [tex]r=\frac{a_{n+1}}{a_n}[/tex]
[tex]\frac{8\left(\frac{1}{4}\right)^{\left(n+1\right)-1}}{8\left(\frac{1}{4}\right)^{n-1}}=\frac{1}{4}[/tex]
The ratio of all the adjacent terms is same and equal to [tex]r=\frac{1}{4}[/tex]
Geometric sequence sum formula: [tex]a_1\frac{1-r^n}{1-r}[/tex]
[tex]n=10,\:\space a_1=8,\:\spacer=\frac{1}{4}[/tex]
[tex]=8\cdot \frac{1-\left(\frac{1}{4}\right)^{10}}{1-\frac{1}{4}}\\\\=\frac{349525}{32768}\\\\\approx 10.67[/tex]
