Finding the closed form of an infinite summation

Question

I tried to figure this out by looking at other answers, but mine seems a little more complex and I can't seem to solve it.

$$\sum\limits_{i=0}^{\infty} \frac{3^{i+1}n}{16}4n = 4n\sum\limits_{i=0}^{\infty} \frac{3^{i+1}n}{16} = 4n^2\sum\limits_{i=0}^{\infty} \frac{3^{i+1}}{16}$$

Answer 1

It simplifies further:

$$3^{i+1}=3\cdot3^i$$

Thus,

$$4n^2\sum\frac{3^{i+1}}{16}=\frac{3n^2}4\sum3^i$$

Now notice that

$$\sum3^i=1+3+9+27+\dots$$

Which does not exist, which means there can not be any closed form.