Answer:
4
Step-by-step explanation:
This is an infinite geometric series. This has a sum of [tex]\frac{a}{1-r}[/tex]
Where
a is the first term, and
r is the common ratio (one term divided by the previous term)
Let's figure out the first 2 terms by plugging in n = 1 first and then n = 2 for the series.
First term:
[tex]3(\frac{1}{4})^{n-1}\\=3(\frac{1}{4})^{1-1}\\=3(\frac{1}{4})^0\\=3(1)\\=3[/tex]
Second term:
[tex]3(\frac{1}{4})^{n-1}\\=3(\frac{1}{4})^{2-1}\\=3(\frac{1}{4})^1\\=3(\frac{1}{4})\\=\frac{3}{4}[/tex]
Let's see the common ratio: [tex]\frac{\frac{3}{4}}{3}\\=\frac{3}{4}*\frac{1}{3}\\=\frac{1}{4}[/tex]
Thus we have a = 3 and r = 1/4. Plugging into the formula of the infinite sum, we get:
[tex]s=\frac{a}{1-4}=\frac{3}{1-\frac{1}{4}}=\frac{3}{\frac{3}{4}}=3*\frac{4}{3}=4[/tex]
So, the answer is 4
