![\sum_{1}^{10}4\left(\frac{1}{4}\right)^{n-1}=4\sum_{1}^{10}\left(\frac{1}{4}\right)^{n-1}\\ \\=4\left[\left(\frac{1}{4}\right)^0+\left(\frac{1}{4}\right)^1+\left(\frac{1}{4}\right)^2+\dots+\left(\frac{1}{4}\right)^9\right]\\ \\=4\left[1+\frac{1}{4}+\left(\frac{1}{4}\right)^2+\dots+\left(\frac{1}{4}\right)^9\right]](/tpl/images/0447/7031/2922a.png)
is a geometric series with first element 
, constant ratio 
and total number 
terms of geometric series 
![1+\frac{1}{4}+\dots+\left(\frac{1}{4}\right)^9=\frac{1\left(1-\left(\frac{1}{4}\right)^{10}\right)}{\left(1-\frac{1}{4}\right)}}\\ \\=1.3333\\ \\4\left[1+\frac{1}{4}+\dots+\left(\frac{1}{4}\right)^9\right]\\ \\=4\times 1.3333\\=5.33\\ \\\sum_{1}^{10}4\left(\frac{1}{4}\right)^{n-1}=5.33](/tpl/images/0447/7031/36815.png)
Evaluate and round to the nearest hundredth
...
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