If $$2^{10} + 2^9 \cdot 3^1 + 2^8 \cdot 3^2 + \ldots + 2 \cdot 3^9 + 3^{10} = S - 2^{11}$$, then $$S$$ is equal to:

We wish to evaluate the finite sum

$$2^{10}+2^{9}\cdot3^{1}+2^{8}\cdot3^{2}+\ldots+2\cdot3^{9}+3^{10}$$

and use the given relation

$$2^{10}+2^{9}\cdot3^{1}+2^{8}\cdot3^{2}+\ldots+2\cdot3^{9}+3^{10}=S-2^{11}.$$

First, we rewrite every term of the sum in a single, systematic form. Observe that the exponent of $$2$$ starts at $$10$$ and decreases by $$1$$ in each successive term, while the exponent of $$3$$ starts at $$0$$ and increases by $$1$$ in each term. Writing the general term, we have

$$\text{General term}=2^{10-k}\,3^{k}\quad\text{for }k=0,1,2,\ldots,10.$$

Hence the entire sum is

$$\sum_{k=0}^{10}2^{10-k}3^{k}.$$

Now factor out the largest constant power of $$2$$ to put the sum into the standard geometric-progression format. We note that $$2^{10-k}=2^{10}\,2^{-k},$$ so

$$\sum_{k=0}^{10}2^{10-k}3^{k}=2^{10}\sum_{k=0}^{10}\left(\frac{3}{2}\right)^{k}.$$

We recognise the inner summation as a finite geometric series with first term $$1$$ and common ratio $$\dfrac{3}{2}.$$ For a geometric series with first term $$a$$, common ratio $$r\neq1$$ and number of terms $$n,$$ the sum is

$$S_n=a\,\frac{r^{\,n}-1}{r-1}.$$

In our case $$a=1,\;r=\dfrac{3}{2},$$ and the number of terms is $$n=11$$ because $$k$$ runs from $$0$$ to $$10$$ (inclusive). Substituting into the formula, we get

$$\sum_{k=0}^{10}\left(\frac{3}{2}\right)^{k}=\frac{\left(\dfrac{3}{2}\right)^{11}-1}{\dfrac{3}{2}-1}.$$

The denominator simplifies to

$$\frac{3}{2}-1=\frac{1}{2}.$$

So the geometric sum is

$$\frac{\left(\dfrac{3}{2}\right)^{11}-1}{\dfrac{1}{2}}=2\left(\left(\frac{3}{2}\right)^{11}-1\right).$$

Multiplying by the outside factor $$2^{10},$$ we obtain the original series value:

$$2^{10}\times2\left(\left(\frac{3}{2}\right)^{11}-1\right)=2^{11}\left(\left(\frac{3}{2}\right)^{11}-1\right).$$

Now simplify the bracket. We have

$$\left(\frac{3}{2}\right)^{11}=\frac{3^{11}}{2^{11}},$$

hence

$$2^{11}\left(\frac{3^{11}}{2^{11}}-1\right)=2^{11}\cdot\frac{3^{11}}{2^{11}}-2^{11}=3^{11}-2^{11}.$$

Therefore the entire sum equals $$3^{11}-2^{11}.$$ According to the statement of the problem, this sum also equals $$S-2^{11}.$$ So we set

$$3^{11}-2^{11}=S-2^{11}.$$

Adding $$2^{11}$$ to both sides, we find

$$S=3^{11}.$$

Hence, the correct answer is Option 2.