Find the partial sum of $a_n=\prod_{k=0}^{n}(2\cdot 3^n)$

Question

Is it possible to find a formula for this partial sum $S_n$?

$$S_n=2+2\times6+2\times6\times18+2\times6\times18\times54+2\times6\times18\times54\times162+\cdots+a_n$$

Here, $\displaystyle a_n=\prod_{k=0}^{n}(2\cdot 3^k)$.

Answer 1

By definition, we have

$$S=\sum_{n=0}^N\prod_{k=0}^n2\times3^k$$

Notice that

$$\prod 2r^{a_k}=2^nr^{\sum a_k}$$

Thus,

$$\prod_{k=0}^n2\times3^k=2^{n+1}3^{\sum_{k=0}^nk}=2^{n+1}3^{n(n+1)/2}=2^{n+1}r^{(n+\frac12)^2-\frac14},r=\sqrt3$$

Thus,

$$S=r^{-1/4}\sum_{n=0}^N2^{n+1}r^{(n+\frac12)^2}$$

which is as far as you can simplify.