Inspired by Ikkyu San

First, let's convert our sum into sigma form for convenience's sake:

$11_{3} + 22_{9} + 33_{27} + \ldots + 99_{19683} = \sum_{k=1}^{9} k3^{k}+k = \sum_{k=1}^{9} k3^{k} + \sum_{k=1}^{9} k$

The latter part of the above sum is clearly easy to handle; the former, not so much. We recall that

$\sum_{k=1}^{n} x^{k} = \frac{x^{n+1}-1}{x-1} \tag{1}$

To obtain a closed form for $\displaystyle \sum_{k=1}^{n} kx^{k}$ , we have to enlist the assistance of an unlikely piece of mathematics: calculus. (You could prove this by induction, but calculus seems more intuitive than bashing out a closed form for the sum in the dark and then inducting.)

We differentiate equation 1:

$\sum_{k=1}^{n} kx^{k-1} = \frac{d}{dx} \frac{x^{n+1}-1}{x-1} = \frac{(n+1)x^{n}(x-1)-(x^{n+1}-1)}{(x-1)^{2}}$

To get $\sum kx^{k}$ all I need is to multiply by $x$ and thus we get

$\sum_{k=1}^{n} kx^{k} = x\frac{(n+1)x^{n}(x-1)-(x^{n+1}-1)}{(x-1)^{2}} = x\frac{(nx-n-1)x^{n}+1}{(x-1)^{2}}$

That was some quality magic right there, I reckon. Quality .

We have to deal with $x = 3$ , and so we obtain

$\sum_{k=1}^{n} k3^{k} = \frac{3}{4} \left( (2n-1)3^{n}+1 \right)$

Now, putting it all together:

$\sum_{k=1}^{9} k3^{k} + \sum_{k=1}^{9} k = \frac{3}{4} \left( (2 \times 9-1)3^{9}+1 \right) + \frac{9(9+1)}{2} = \boxed{251004}$