#### To determine

Whether the 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=1∞(35n+2n).

**Result used:**

(1) The geometric series ∑n=1∞arn−1 (or) a+ar+ar2+⋯ is convergent if |r|<1 and its sum is a1−r.

(2) 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:**

The given series can be written as follows,

∑n=1∞(35n+2n)=∑n=1∞35n+∑n=1∞2n=3⋅∑n=1∞15n+2⋅∑n=1∞1n

=3⋅∑n=1∞(15)n+2⋅∑n=1∞1n (1)

Here, ∑n=1∞1n is harmonic series and it is divergent.

Note that, a nonzero multiple of a divergent series is also divergent.

Therefore, the series 2⋅∑n=1∞1n is divergent.

Consider the series, ∑n=1∞(15)n=(15)+(15)2+(15)3+⋯.

Clearly, it is geometric series with the first term of the series, a=15 and the common ratio of the series is,

r=(15)215=15

The absolute value of *r* is,

|r|=|15|=15=0.2<1

Since |r|<1 and by the Result (1) stated above, the series is convergent.

Thus, the series ∑n=1∞(15)n is convergent.

Note that, a nonzero multiple of a convergent series is also convergent.

Therefore, the series 3⋅∑n=1∞(15)n is convergent.

Note that, the sum of a convergent and a divergent series is divergent.

Since the series 3⋅∑n=1∞(15)n is convergent and 2⋅∑n=1∞1n is divergent, then their sum ∑n=1∞(2e)n+∑n=1∞(4e)n is divergent.

From equation (1), the series ∑n=1∞(35n+2n) is divergent.

Therefore, the series is divergent.