I'm trying to solve the following triple summation, but I'm not sure this is the best way to solve it, here's what I tried: $$\sum_{k=3}^{n} \sum_{i=2}^{k-1}\sum_{j=0}^{i} 3^{i+j+k}$$
$n \in \mathbb{N}, n\geq 3$ : $$\sum_{k=3}^{n} \sum_{i=2}^{k-1}\sum_{j=0}^{i} 3^{i+j+k}$$ $$=\sum_{k=3}^{n} 3^{k} \sum_{i=2}^{k-1}3^{i} \sum_{j=0}^{i} 3^{j}$$ $$=\sum_{k=3}^{n} 3^{k} \sum_{i=2}^{k-1}3^{i} \frac{1-3^{i+1}}{1-3}$$ $$=\sum_{k=3}^{n} \frac{3^{k}}{2} \sum_{i=2}^{k-1}(-3^{i}+3^{2i+1})$$ $$=\sum_{k=3}^{n} \frac{3^{k}}{2} \big((-1)\sum_{i=2}^{k-1}3^{i}+ 3\sum_{i=2}^{k-1}9^{i})\big)$$ $$=\sum_{k=3}^{n} \frac{3^{k}}{2} \big((-1)(\frac{1-3^{k-1+1}}{1-3}-3^0-3^1)+3(\frac{1-9^{k-1+1}}{1-9}-9^0-9^1)\big)$$ $$=\sum_{k=3}^{n} \frac{3^{k}}{2} \big(\frac{1}{2}-\frac{3^k}{2}+4-\frac{3}{8}+\frac{3}{8}9^k-30\big)$$ $$=\sum_{k=3}^{n} \frac{3^{k}}{2} \big(\frac{-207}{8}-\frac{3^k}{2}+\frac{3}{8}9^k\big)$$ $$=\frac{-207}{16}\sum_{k=3}^{n} 3^{k} - \frac{1}{4}\sum_{k=3}^{n} 9^k+\frac{3}{16}\sum_{k=3}^{n}27^k$$ $$=\frac{-207}{16}(\frac{1-3^{n+1}}{1-3}-3^0-3^1-3^2) - \frac{1}{4}(\frac{1-9^{n+1}}{1-9}-9^0-9^1-9^2)+\frac{3}{16}(\frac{1-27^{n+1}}{1-27}-27^0-27^1-27^2)$$ $$=\frac{207}{32} - \frac{207}{32} 3^{n+1} + \frac{2691}{16}+ \frac{1}{32}-\frac{1}{32}9^{n+1} +\frac{91}{4}-\frac{3}{416}+\frac{3}{416}27^{n+1}-\frac{2271}{16}$$ $$=\frac{23085}{416} - \frac{207}{32} 3^{n+1}-\frac{1}{32}9^{n+1}+\frac{3}{416}27^{n+1}$$
Your computation is correct, but it is a little tedious. Some simplification is possible if we are strategic. First, let's look at the innermost double sum: $$\sum_{i=2}^{k-1} \sum_{j=0}^i 3^{i+j+k}.$$ If we let $s = i-2$, then this becomes $$\sum_{s=0}^{k-3} \sum_{j=0}^{s+2} 3^{s+2+j+k} = 3^2 \sum_{s=0}^{k-3} \sum_{j=0}^{s+2} 3^{s+j+k}.$$ Now the outermost sum goes from $k = 3$ to $k = n$, so let $m = k-3$: $$3^2 \sum_{k=3}^n \sum_{s=0}^{k-3} \sum_{j=0}^{s+2} 3^{s+j+k} = 3^2 \sum_{m = 0}^{n-3} \sum_{s=0}^m \sum_{j=0}^{s+2} 3^{s+j+m+3} = 3^5 \sum_{m=0}^{n-3} \sum_{s=0}^m \sum_{j=0}^{s+2} 3^{s+j+m}.$$ Now all of the lower indices of summation start at $0$, which makes things a bit nicer. We now have $$\begin{align} \sum_{m=0}^{n-3} \sum_{s=0}^m \sum_{j=0}^{s+2} 3^{s+j+m} &= \sum_{m=0}^{n-3} 3^m \sum_{s=0}^{m} 3^s \sum_{j=0}^{s+2} 3^j \\ &= \sum_{m=0}^{n-3} 3^m \sum_{s=0}^m 3^s \frac{3^{s+3} - 1}{2} \\ &= \frac{1}{2} \sum_{m=0}^{n-3} 3^m \sum_{s=0}^{m} (3^{2s+3} - 3^s) \\ &= \frac{1}{2} \sum_{m=0}^{n-3} 3^m \left( 3^3 \frac{3^{2(m+1)} - 1}{3^2 - 1} - \frac{3^{m+1} - 1}{3-1} \right) \\ &= \frac{1}{16} \sum_{m=0}^{n-3} 3^m(3^{2m+5} - 3^3 - 4 \cdot 3^{m+1} + 4) \\ &= \frac{1}{16} \sum_{m=0}^{n-3} 3^{3m+5} - 4 \cdot 3^{2m+1} - 23\cdot 3^m \\ &= \frac{1}{16} \left(3^5 \frac{3^{3(n-2)}-1}{3^3-1} - 12 \frac{3^{2(n-2)}-1}{3^2-1} - 23 \frac{3^{n-2} - 1}{3-1} \right) \\ &= \frac{1}{16} \left(\frac{3^{3n-1}}{26} - \frac{3^{2n-3}}{2} - \frac{23\cdot 3^{n-2}}{2} - \frac{243}{26} + \frac{3}{2} + \frac{23}{2}\right) \\ &= \frac{1}{16} \left(\frac{3^{3n}}{78} - \frac{3^{2n}}{54} - \frac{23}{18}3^n + \frac{95}{26} \right). \end{align}$$
Then the final sum is simply $3^5$ times this result.