Suppose $$\left(\frac 12+\frac 13+\cdots +\frac 1{149}\right)$$ as $$A$$, $$\left(\frac 12+\frac 13+\cdots +\frac 1{150}\right)$$ as $$B$$.  $$(1+A)\times B-(1+B)\times A=B+AB-A-AB=B-A=\frac 1{150}$$. 
Calculate:  $$\left(1+\frac 12+\frac 13+\cdots +\frac 1{149}\right)\times \left(\frac 12+\frac 13+\cdots +\frac 1{149}+\frac 1{150}\right)$$$$-\left(1+\frac 12+\frac 13+\cdots +\frac 1{149}+\frac 1{150}\right)\times \left(\frac 12+\frac 13+\cdots +\frac 1{149}\right)=$$~\uline{~~~~~~~~~~}~. 
Options:
(A) $$1$$ 
(B) $\dfrac{1}{149}$ 
(C) $\dfrac{149}{150}$ 
(D) $\dfrac{1}{150}$ 
(E) $$150$$ 