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For what value of x is \(1+x+x^2+x^3+...=4\)?

 Jun 16, 2023
 #1
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Let's start by writing the equation as a geometric series:

1 + x + x^2 + x^3 + ... = 4

We can then factor out a 1 from the left-hand side:

1(1 + x + x^2 + x^3 + ...) = 4

We can then divide both sides by 1 - x to get:

1/(1 - x) = 4

This is an equation of the form a/b = c, where a, b, and c are all real numbers. The solution to this equation is:

x = -b/a

In this case, a = 1, b = 1, and c = 4. Plugging these values into the solution formula, we get:

x = -b/a = -1/1 = -1

However, this is not the solution we are looking for. The series 1 + x + x^2 + x^3 + ... diverges when x = -1. This is because the absolute value of the common ratio (x) is greater than 1.

We need to find a solution that makes the series converge. To do this, we need to find a value of x for which the absolute value of the common ratio is less than 1. The only real number x for which this is true is x = -1/3.

When x = -1/3, the common ratio is x = -1/3. The absolute value of this is less than 1, so the series converges. Therefore, the solution to 1 + x + x^2 + x^3 + ... = 4 is x = -1/3.

 Jun 16, 2023
 #2
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Like so:

 

 Jun 16, 2023

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