Solving for x in an Infinite Geometric Series

The question asks us to find the relationship between \(x\) and \(y\), where \(y\) is defined as an infinite series:

\[y = x + x^2 + x^3 + \dots\]

This is an example of an infinite geometric series. A geometric series is a series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

Identifying the Geometric Series Parameters

In the given series \(y = x + x^2 + x^3 + \dots\):

• The first term, \(a\), is \(x\).
• The common ratio, \(r\), is found by dividing any term by the previous term. For example, \(x^2 / x = x\), or \(x^3 / x^2 = x\). So, the common ratio \(r\) is \(x\).

Sum of an Infinite Geometric Series Formula

An infinite geometric series converges (has a finite sum) if and only if the absolute value of the common ratio is less than 1, i.e., \(|r| < 1\). The question states that \(x < 1\). Assuming \(x\) is such that \(|x| < 1\) (which is required for convergence), the sum \(S\) of an infinite geometric series is given by the formula:

\[S = \frac{a}{{1 - r}}\]

Applying the Formula to the Given Series

Here, the sum is \(y\), the first term \(a = x\), and the common ratio \(r = x\). Substituting these into the formula, we get:

\[y = \frac{x}{{1 - x}}\]

Solving for x in Terms of y

Now we need to rearrange this equation to express \(x\) in terms of \(y\). The steps are as follows:

Start with the equation:

\[y = \frac{x}{{1 - x}}\]

Multiply both sides by \((1 - x)\) to clear the denominator:

\[y(1 - x) = x\]

Distribute \(y\) on the left side:

\[y - yx = x\]

Move the term containing \(x\) from the left side (\(-yx\)) to the right side by adding \(yx\) to both sides:

\[y = x + yx\]

Factor out \(x\) from the terms on the right side:

\[y = x(1 + y)\]

Finally, divide both sides by \((1 + y)\) to isolate \(x\):

\[x = \frac{y}{{1 + y}}\]

Comparing with the Options

The expression we found for \(x\) is \(\frac{y}{{1 + y}}\). Let's compare this with the given options:

• Option 1: \(x = \frac{y}{{1 + y}}\)
• Option 2: \(x = \frac{y}{{1 - y}}\)
• Option 3: \(x = \frac{{1 + y}}{y}\)
• Option 4: \(x = \frac{{1 - y}}{y}\)

Our result matches Option 1.

Thus, if \(y = x + x^2 + x^3 + \dots\) up to infinite terms where \(x < 1\) (and \(|x|<1\)), then \(x = \frac{y}{{1 + y}}\).

Revision Table: Key Concepts

Additional Information: Conditions for Convergence

The sum of an infinite geometric series \(a + ar + ar^2 + \dots\) converges to a finite value if and only if the absolute value of the common ratio \(r\) is less than 1, i.e., \(|r| < 1\). If \(|r| \ge 1\), the terms of the series do not approach zero, and the sum diverges (does not have a finite value).

In this problem, the common ratio is \(x\). The question states \(x < 1\). For convergence of the series \(x + x^2 + x^3 + \dots\), we strictly need \(|x| < 1\). The given condition \(x < 1\) covers cases where \(x\) is between -1 and 1 (exclusive of 1) and also cases where \(x\) is less than or equal to -1. However, the sum formula \(S = \frac{a}{{1 - r}}\) is only valid when \(|r| < 1\). Thus, the problem implicitly assumes that \(x\) is a value for which the series converges, meaning \(|x| < 1\).

The derived formula \(x = \frac{y}{{1 + y}}\) assumes this convergence holds.