To determine

Whether the geometric series is convergent or divergent and obtain the sum if the series is convergent.

Answer

The series is divergent.

Explanation

Given:

The series is ∑n=0∞3n+1(−2)n .

Result used:

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

Calculation:

Consider the given series ∑n=0∞3n+1(−2)n .

∑n=0∞3n+1(−2)n=30+1(−2)0+31+1(−2)1+32+1(−2)2+33+1(−2)3+⋯=311+32−2+334+34−8+⋯=3−92+274−818+⋯

Here, the first term of the series is a = 3 and the common ratio of the series is,

r=Second termFirst term=−923=−92⋅13=−32

The absolute value of r is,

|r|=|−32|=32=1.5>1

Since |r|>1 and by using the Result stated above, the series is divergent.

Therefore, the series is divergent.