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# geometric series

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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