Find closed formula by changing order of summation: $\sum_{i=1}^ni3^i$
Here is a rather detailed elaboration which might be useful.
We obtain \begin{align*} \color{blue}{\sum_{i=1}^ni3^i}&=\sum_{i=1}^n\left(\sum_{k=1}^i 1\right)3^i\tag{1}\\ &=\sum_{i=1}^n\sum_{k=1}^i 3^i =\sum_{1\leq k\leq i\leq n}3^i =\sum_{k=1}^n\sum_{i=k}^n3^i\tag{2}\\ &=\sum_{k=1}^n\sum_{i=0}^{n-k}3^{i+k}\tag{3}\\ &=\sum_{k=1}^n3^k\cdot\frac{3^{n-k+1}-1}{3-1}\tag{4}\\ &=\frac{1}{2}\sum_{k=1}^n\left(3^{n+1}-3^k\right)\tag{5}\\ &=\frac{n}{2}3^{n+1}-\frac{1}{2}\sum_{k=1}^n3^k\tag{6}\\ &=\frac{n}{2}3^{n+1}-\frac{1}{2}\cdot\left(\frac{3^{n+1}-1}{3-1}-1\right)\tag{7}\\ &=\frac{n}{2}3^{n+1}-\frac{1}{4}3^{n+1}+\frac{3}{4}\tag{8}\\ &\color{blue}{=\frac{n}{4}(2n-1)3^{n+1}+\frac{3}{4}}\tag{9} \end{align*}
Comment:
In (1) we represent the factor $i$ as sum.
In (2) we multiply out in the left-hand sum and write the index range somewhat more conveniently in the middle sum. We exchange the sums in the right-hand double-sum.
In (3) we shift the index of the inner sum to start from $i=0$.
In (4) we apply the finite geometric summation formula.
In (5) we do some simplifications.
In (6) we multiply out and do some simplifications.
In (7) we apply the finite geometric summation formula again.
In (8) and (9) we do some more simplifications.
Just use a sum of the geometric series.
It's $$\sum_{i=1}^n\frac{3^k(3^{n-k+1}-1)}{3-1}=\frac{n3^{n+1}}{2}-\frac{1}{2}\frac{3(3^n-1)}{2}=\frac{2n3^{n+1}-3^{n+1}+3}{4}.$$
Because the other answer already discusses about computing $\sum_{k=1}^n \sum_{i=k}^n 3^i$, here's another way to compute $\sum_{i=1}^n i3^i$
\begin{eqnarray} S & = & \sum_{i=1}^n i 3^i \\ & = & 1.3^1 + 2.3^2 + 3.3^3 + \cdots + n.3^n \\ 3S & = & \space\space\space\space\space\space\space\space\space\space\space\space 1.3^2 + 2.3^3 + \cdots + (n-1).3^n + n.3^{n+1} \\ \end{eqnarray}
Therefore, on subtracting, we get $$S - 3S = (3^1 + 3^2 + 3^3 + \cdots + 3^n) - n.3^{n+1}$$ $$\implies -2S = 3\left(\frac{3^n-1}{2}\right) - n.3^{n+1} = \frac{(1-2n)3^{n+1}-3}{2}$$ $$\implies S = \frac{(2n-1)3^{n+1}+3}{4}$$