If $$1 + \left(\frac{1}{2}\right) + \left(\frac{1}{3}\right) + … + \left(\frac{1}{20}\right) = k$$, then what is the value of $$\left(\frac{1}{4}\right) + \left(\frac{1}{6}\right) + \left(\frac{1}{8}\right) + … + \left(\frac{1}{40}\right)$$?

$$1+\left(\frac{1}{2}\right)+\left(\frac{1}{3}\right)+\dots+\left(\frac{1}{20}\right)=k\ .$$

or, $$\left(\frac{1}{2}\right)+\left(\frac{1}{3}\right)+\dots+\left(\frac{1}{20}\right)=k-1\ .$$

So, $$\left(\frac{1}{4}\right) + \left(\frac{1}{6}\right) + \left(\frac{1}{8}\right) + … + \left(\frac{1}{40}\right)$$

$$=\frac{1}{2}\left(\left(\frac{1}{2}\right)+\left(\frac{1}{3}\right)+\dots+\left(\frac{1}{20}\right)\right)$$

$$=\frac{1}{2}\left(k-1\right)\ .$$

C is correct choice.