#### 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 (−1,5) and the sum of the series is 35−x.

#### Explanation

**Given:**

The series is ∑n=0∞(x−2)n3n.

**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∞(x−2)n3n=∑n=0∞(x−23)n=(x−23)0+(x−23)1+(x−23)2+(x−23)3+⋯=1+(x−23)+(x−23)2+(x−23)3+⋯

Clearly, it is geometric series with first term of the series is a=1 and common ratio is r=x−23.

Use the Result stated above, the geometric series ∑n=0∞(x−2)n3n is converges if |r|<1.

|x−23|<1|x−2|<3−3<x−2<3−3+2<x−2+2<3+2

That is, −1<x<5.

Thus, the series is converges if x∈(−1,5).

Therefore, the interval of convergence is (−1,5).

Obtain the sum of the geometric series.

Since a=1 and r=x−23, the sum of the series is computed as follows.

∑n=0∞(x−2)n3n=11−(x−23)=13−x+23=15−x3=35−x

Therefore, the sum of the series is 35−x.