#1**0 **

Factor b:

\(b(\frac{1}{7}+\frac{b}{7^2}+\frac{b^2}{7^3}+...)=1\)

Now, what is inside the bracket is a geometric series with a common ratio, \(r=\frac{b}{7}\)

The formula is:

\(Sum=\frac{1}{1-r}=\frac{1}{1-\frac{b}{7}}=\frac{7}{7-b}\) (But, r has to be less than 1).

So:

\(b(\frac{7}{7-b})=1\), can you solve it from here?

Guest Jul 19, 2022

#2**0 **

Because it is converging to 1, therefore:

sumfor(n, 1, infinity, ((b)^n) / (7^n))==1

**b ==7 / 2 ==3.5**

Guest Jul 19, 2022

#3**0 **

The formula for the sum is \({s \over {1 - r}} \), where s is the starting term, and r is the common ratio.

The starting term is \({b \over 7} \), and the common term is \({b \over 7}\). This means that we have the equation \({{b \over 7} \over {1 - {b \over 7}}} = 1\).

Solving for b, we find \(b = {7 \over 2} = \color{brown}\boxed{3.5}\)

BuilderBoi Jul 19, 2022