It's the sum of a geometric sequence.
Let's rewrite it a bit:
[tex]\displaystyle\\\sum_{n=1}^{10}8\left(\dfrac{1}{4}\right)^{n-1}=8\sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n\cdot \left(\dfrac{1}{4}\right)^{-1}=8\sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n\cdot 4=32\sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n[/tex]
And now let's calculate this sum [tex]\displaystyle \sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n[/tex]:
[tex]S_n=\dfrac{a(1-r^n)}{1-r}\\\\a=\dfrac{1}{4}\\n=10\\r=\dfrac{1}{4}\\\\S_{10}=\dfrac{\dfrac{1}{4}\cdot\left(1-\left(\dfrac{1}{4}\right)^{10}\right)}{1-\dfrac{1}{4}}=\dfrac{\dfrac{1}{4}\cdot\left(1-\dfrac{1}{1048576}\right)}{\dfrac{3}{4}}=\dfrac{\dfrac{1048575}{1048576}}{3}=\dfrac{1048575}{3145728}=\\=\dfrac{349525}{1048576}[/tex]
Now let's calculate the initial sum:
[tex]\displaystyle 32\cdot \sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n=32\cdot \dfrac{349525}{1048576}=\dfrac{349525}{32768}\approx10.67[/tex]
