This summation where $n \in \mathbb{N}$:
$a(n)=\sum_{i=1}^{n} 2^{i-1}3^{n-i}$
...gives rise to this sequence for $a(n)$:
1,5,19,65,211...
...which is OEIS A001047.
The OEIS entry has the formula as $a(n)=3^n-2^n$.
My question is how to derive $\sum_{i=1}^{n} 2^{i-1}3^{n-i}=3^n-2^n$
Hint: write it as $$ \sum_{i=1}^n 3^n 2^{-1} (2/3)^i $$ and calculate the geometric sum
$$ \sum_{i=1}^n 2^{i-1}3^{n-i} = \frac{3^n}{2}\sum_{i=1}^n (\frac{2}{3})^i = \frac{3^n}{2} \frac{2}{3}\frac{1-(\frac{2}{3})^n}{1-\frac{2}{3}} = \frac{1}{2}\frac{2}{3} \frac{3^n-2^n}{\frac{1}{3}} = \frac{3}{2}\frac{2}{3} (3^n-2^n) = 3^n -2^n$$
Note that $$(3-2)\sum_{i=1}^n 2^{i-1}3^{n-i}=\sum_{i=1}^n(2^{i-1}3^{n-i+1}-2^{i} 3^{n-i})$$ and the sum collapses to $3^n-2^n$.
Hint
$$\sum_{i=1}^n2^{i-1}3^{n-i}=\frac{3^n}{2}\sum_{i=1}^n\left(\frac{2}{3}\right)^i,$$ and you should recognize this last famous sum.
Another approach, based on the generalized case here. This does not require full expansion of the original summation.
$$\begin{align} \sum_{i=1}^n 2^{i-1}3^{n-i} &=\sum_{i=1}^n 2^{i-1}(1+2)^{n-i}\\ &=\sum_{i=1}^n 2^{i-1}\sum_{j=0}^{n-i}\binom {n-i}j2^j\\ &=\sum_{i=1}^n \sum_{j=0}^{n-i}\binom {n-i}j2^{i+j-1}\\ &=\sum_{i=1}^n\sum_{k=i-1}^{n-1}\binom {n-i}{i-i+1}2^k &&(k=i+j-1)\\ &=\sum_{i=1}^n\sum_{k=i}^{n}\binom {n-i}{k-i}2^{k-1}\\ &=\sum_{k=1}^n2^{k-1}\sum_{i=1}^k \binom {n-i}{n-k}\\ &=\sum_{k=1}^n 2^{k-1} \sum_{r=n-k}^{n-1}\binom r{n-k} &&(r=n-i)\\ &=\sum_{k=1}^n2^{k-1}\binom n{n-k+1}\\ &=\sum_{k=1}^n 2^{k-1}\binom n{k-1}\\ &=\sum_{k=0}^{n-1} \binom nk 2^k\\ &=\sum_{k=0}^n \binom nk 2^k-2^n\\ &=\color{red}{3^n-2^n} \end{align}$$
