Answer:
[tex]5.33[/tex]
Step-by-step explanation:
[tex]\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][/tex]
[tex]1+\frac{1}{4}+\dots+\left(\frac{1}{4}\right)^9[/tex] is a geometric series with first element [tex](a)=1[/tex], constant ratio [tex](r)=\frac{1}{4}[/tex] and total number [tex](n)=10[/tex]
sum of [tex]n[/tex] terms of geometric series [tex]=\frac{a(1-r^{n})}{1-r}[/tex]
[tex]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[/tex]
