The sum $$\sum_{k=1}^{20} k \cdot \frac{1}{2^k}$$ is equal to:

We wish to evaluate

$$S=\sum_{k=1}^{20}k\left(\frac12\right)^k.$$

To handle the term $$k$$, we first recall the well-known finite geometric-series formula

$$\sum_{k=0}^{n}x^{k}=\frac{1-x^{\,n+1}}{1-x}\qquad\text{for }x\neq1.$$

Now we differentiate both sides with respect to $$x$$. Differentiating term-by-term on the left gives

$$\frac{d}{dx}\left(\sum_{k=0}^{n}x^{k}\right)=\sum_{k=0}^{n}k\,x^{\,k-1}.$$

Differentiating the right-hand side by the quotient rule (or by first writing it as $$(1-x^{\,n+1})(1-x)^{-1}$$) yields

$$\frac{d}{dx}\left(\frac{1-x^{\,n+1}}{1-x}\right)= \frac{-(n+1)x^{\,n}(1-x)+(1-x^{\,n+1})}{(1-x)^2}.$$ A little simplification changes its numerator to $$-(n+1)x^{\,n}(1-x)+(1-x^{\,n+1}) =-(n+1)x^{\,n}+ (n+1)x^{\,n+1}+1-x^{\,n+1} =1-(n+1)x^{\,n}+n\,x^{\,n+1}.$$ Hence we have the identity

$$\sum_{k=0}^{n}k\,x^{\,k-1}=\frac{1-(n+1)x^{\,n}+n\,x^{\,n+1}}{(1-x)^2}.$$

Multiplying both sides by $$x$$ converts $$x^{\,k-1}$$ into $$x^{\,k}$$ and gives the standard result

$$\sum_{k=0}^{n}k\,x^{\,k}=x\;\frac{1-(n+1)x^{\,n}+n\,x^{\,n+1}}{(1-x)^2}.$$ Because the $$k=0$$ term is $$0$$, we can also write

$$\sum_{k=1}^{n}k\,x^{\,k}=x\;\frac{1-(n+1)x^{\,n}+n\,x^{\,n+1}}{(1-x)^2}.$$

We now set $$x=\dfrac12$$ and $$n=20$$ for our problem. First compute the denominator:

$$1-x=\;1-\frac12=\frac12,\qquad (1-x)^2=\left(\frac12\right)^2=\frac14.$$

Substituting $$x=\dfrac12$$ into the general formula gives

$$S=\frac12\;\frac{\,1-(20+1)\left(\dfrac12\right)^{20}+20\left(\dfrac12\right)^{21}\!}{\left(\dfrac12\right)^2}.$$ Because dividing by $$\left(\dfrac12\right)^2=\dfrac14$$ is the same as multiplying by $$4$$, we obtain

$$S= \frac12 \times 4 \,\Bigl[1-(21)\,2^{-20}+20\,2^{-21}\Bigr].$$

Simplifying the factor $$\dfrac12\times4$$ first:

$$\frac12\times4=2,$$ so

$$S=2\Bigl[1-21\cdot2^{-20}+20\cdot2^{-21}\Bigr].$$

Now we separate the terms:

$$S=2-2\cdot21\cdot2^{-20}+2\cdot20\cdot2^{-21}.$$ Observe that $$2\cdot21\cdot2^{-20}=\frac{42}{2^{20}}=\frac{21}{2^{19}},\qquad 2\cdot20\cdot2^{-21}=\frac{40}{2^{21}}=\frac{20}{2^{20}}.$$

Hence

$$S=2-\frac{21}{2^{19}}+\frac{20}{2^{20}}.$$

To combine the two fractional terms we write the first with denominator $$2^{20}$$:

$$\frac{21}{2^{19}}=\frac{42}{2^{20}},$$ so $$S=2-\frac{42}{2^{20}}+\frac{20}{2^{20}} =2-\frac{42-20}{2^{20}} =2-\frac{22}{2^{20}}.$$

Finally we notice that $$\dfrac{22}{2^{20}}=\dfrac{11}{2^{19}}$$, because multiplying numerator and denominator by $$2$$ does not change the value. Therefore

$$S=2-\frac{11}{2^{19}}.$$

Among the given options, this matches Option D.

Hence, the correct answer is Option D.