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Find the value of b given that
b/7+b^2/7^2+b^3/7^3.....=1

 Jul 19, 2022
 #1
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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?

 Jul 19, 2022
 #2
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Because it is converging to 1, therefore:

 

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

 

b ==7 / 2 ==3.5

 Jul 19, 2022
 #3
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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 \(b = {7 \over 2} = \color{brown}\boxed{3.5}\)

 Jul 19, 2022

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