The sum of the first 20 terms of the series $$1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \ldots$$ is:

We observe the terms one by one:

$$1,\;\; \frac32,\;\; \frac74,\;\; \frac{15}8,\;\; \frac{31}{16},\ldots$$

Let us try to write the general term. In the numerators we have $$1,3,7,15,31,\ldots$$ and we note that

$$1=2^{1}-1,\; 3=2^{2}-1,\; 7=2^{3}-1,\;15=2^{4}-1,\ldots$$

In the denominators we have $$1,2,4,8,16,\ldots=2^{0},2^{1},2^{2},2^{3},2^{4},\ldots$$

Hence the $$n^{\text{th}}$$ term, say $$T_n$$, is

$$T_n=\frac{2^{\,n}-1}{2^{\,n-1}}.$$

Now we simplify this term:

$$T_n=\frac{2^{\,n}}{2^{\,n-1}}-\frac{1}{2^{\,n-1}} =2-\frac1{2^{\,n-1}}.$$

So every term can be seen as the difference of a constant $$2$$ and a power of $$\tfrac12$$.

We need the sum of the first 20 terms. Let

$$S_{20}=\sum_{n=1}^{20}T_n =\sum_{n=1}^{20}\left(2-\frac1{2^{\,n-1}}\right).$$

We split this sum into two separate summations:

$$S_{20} =\sum_{n=1}^{20}2-\sum_{n=1}^{20}\frac1{2^{\,n-1}}.$$

The first summation is easy:

$$\sum_{n=1}^{20}2=2\times20=40.$$

For the second summation we notice a finite geometric series with first term $$a=1$$ and common ratio $$r=\tfrac12$$.

Formula for the sum of the first $$N$$ terms of a G.P. is

$$S_N=\frac{a(1-r^{\,N})}{1-r}.$$

Putting $$N=20$$, $$a=1$$ and $$r=\tfrac12$$, we get

$$\sum_{n=1}^{20}\frac1{2^{\,n-1}} =\frac{1\left(1-\left(\tfrac12\right)^{20}\right)}{1-\tfrac12} =\frac{1-\tfrac1{2^{20}}}{\tfrac12} =2\Bigl(1-\tfrac1{2^{20}}\Bigr) =2-\frac{2}{2^{20}} =2-\frac1{2^{19}}.$$

Substituting both partial sums back, we find

$$S_{20}=40-\left(2-\frac1{2^{19}}\right) =40-2+\frac1{2^{19}} =38+\frac1{2^{19}}.$$

Hence, the correct answer is Option C.