Question

S=a1/1-r

Answer 1

Answer:

4

Step-by-step explanation:

This is an infinite geometric series. This has a sum of  $\frac{a}{1-r}$

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:

$3(\frac{1}{4})^{n-1}\\=3(\frac{1}{4})^{1-1}\\=3(\frac{1}{4})^0\\=3(1)\\=3$

Second term:

$3(\frac{1}{4})^{n-1}\\=3(\frac{1}{4})^{2-1}\\=3(\frac{1}{4})^1\\=3(\frac{1}{4})\\=\frac{3}{4}$

Let's see the common ratio:  $\frac{\frac{3}{4}}{3}\\=\frac{3}{4}*\frac{1}{3}\\=\frac{1}{4}$

Thus we have a = 3 and r = 1/4. Plugging into the formula of the infinite sum, we get:

$s=\frac{a}{1-4}=\frac{3}{1-\frac{1}{4}}=\frac{3}{\frac{3}{4}}=3*\frac{4}{3}=4$

So, the answer is 4