The series . ` (2x)/(x + 3) + ((2x)/(x + 3))^(2) + ((2x)/(x + 3))^(3) + ...oo`
have a definite sum when

Options

A. ` - 1 lt x lt 3 `
B. `0 lt x lt 1 `
C. `x = 0 `
D. `x = - 3 `

Answer

To find the values of \( x \) for which the series

\[ \frac{2x}{x + 3} + \left(\frac{2x}{x + 3}\right)^{2} + \left(\frac{2x}{x + 3}\right)^{3} + \ldots \]

has a definite sum, we recognize that this is an infinite geometric series.

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the series is:

\[ a = \frac{2x}{x + 3} \]

The common ratio \( r \) is also:

\[ r = \frac{2x}{x + 3} \]

### Step 2: Condition for convergence of the series
For the infinite geometric series to converge, the absolute value of the common ratio must be less than 1:

\[ |r| < 1 \]

This means:

\[ \left|\frac{2x}{x + 3}\right| < 1 \]

### Step 3: Solve the inequality
We will break this absolute value inequality into two cases.

#### Case 1:
\[ \frac{2x}{x + 3} < 1 \]

Multiplying both sides by \( x + 3 \) (noting that we need to consider the sign of \( x + 3 \)) gives:

\[ 2x < x + 3 \]

Rearranging this gives:

\[ x < 3 \]

#### Case 2:
\[ \frac{2x}{x + 3} > -1 \]

Again, multiplying both sides by \( x + 3 \) (considering the sign) gives:

\[ 2x > -x - 3 \]

Rearranging this gives:

\[ 3x > -3 \quad \Rightarrow \quad x > -1 \]

### Step 4: Combine the results
From the two cases, we have:

1. \( x < 3 \)
2. \( x > -1 \)

Thus, combining these results, we find:

\[ -1 < x < 3 \]

### Conclusion
The series has a definite sum when \( x \) lies in the interval:

\[ (-1, 3) \]

### Final Answer
The answer is that the series has a definite sum when \( x \) lies between -1 and 3, which corresponds to option A.

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