#### To determine

**To find:** The values of *x* if the given series is convergent and obtain the sum of the series.

#### Answer

The interval of convergence is (−∞,−2)∪(2,∞) and the sum of the series is xx−2.

#### Explanation

**Given:**

The series is ∑n=0∞2nxn.

**Result used:**

The geometric series ∑n=1∞arn−1 (or) a+ar+ar2+⋯ is converges if |r|<1 and its sum is a1−r, where *a* is the first term and *r* is the common ratio of the series.

**Calculation:**

Obtain the value of *x* (the interval of converges).

The given series can be expressed as follows,

∑n=0∞2nxn=∑n=0∞(2x)n=(2x)0+(2x)1+(2x)2+(2x)3+⋯=1+(2x)+(2x)2+(2x)3+⋯

Clearly, it is geometric series with first term of the series is a=1 and common ratio is r=2x.

Use the Result stated above, the geometric series ∑n=0∞2nxn is converges if |r|<1.

|2x|<1|2||x|<1 [By properties of modulus, |ab|=|a||b|]2|x|<1

Take reciprocal on both sides,

|x|2>1

Multiply the above inequality by 2,

|x|2⋅2>1⋅2|x|>2−x>2 (or) x>2

That is, x<−2 (or) x>2.

Thus, the series is converges if x∈(−∞,−2)∪(2,∞).

Therefore, the interval of convergence is (−∞,−2)∪(2,∞).

Obtain the sum of the geometric series.

Since a=1 and r=2x.

∑n=0∞2nxn=11−2x=1xx−2x=1x−2x=xx−2

Therefore, the sum of the series is xx−2.