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?
Because it is converging to 1, therefore:
sumfor(n, 1, infinity, ((b)^n) / (7^n))==1
b ==7 / 2 ==3.5
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 \(b = {7 \over 2} = \color{brown}\boxed{3.5}\)