Question

Can anyone help me solve this? Please and thank you!

Answer 1

It's the sum of a geometric sequence.

Let's rewrite it a bit:

$\displaystyle\\\sum_{n=1}^{10}8\left(\dfrac{1}{4}\right)^{n-1}=8\sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n\cdot \left(\dfrac{1}{4}\right)^{-1}=8\sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n\cdot 4=32\sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n$

And now let's calculate this sum $\displaystyle \sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n$:

$S_n=\dfrac{a(1-r^n)}{1-r}\\\\a=\dfrac{1}{4}\\n=10\\r=\dfrac{1}{4}\\\\S_{10}=\dfrac{\dfrac{1}{4}\cdot\left(1-\left(\dfrac{1}{4}\right)^{10}\right)}{1-\dfrac{1}{4}}=\dfrac{\dfrac{1}{4}\cdot\left(1-\dfrac{1}{1048576}\right)}{\dfrac{3}{4}}=\dfrac{\dfrac{1048575}{1048576}}{3}=\dfrac{1048575}{3145728}=\\=\dfrac{349525}{1048576}$

Now let's calculate the initial sum:

$\displaystyle 32\cdot \sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n=32\cdot \dfrac{349525}{1048576}=\dfrac{349525}{32768}\approx10.67$