That would take a long time for us to calculate!!Instead,let's try something else.

Using the theory of telescopic sum, we have $\frac{1}{k(k+n)}$ , where we have: $\frac{1}{k}$ - $\frac{1}{k + n}$ , but if $n > 1$ , we will need to divide the numerator by $n$ . Starting: For $A$ we have: $\frac{2}{k(k+2)}$ = $\frac{2}{1}$ - $\frac{2}{3}$ + $\frac{2}{ 3 }$ + ... + $\frac{2}{9}$ - $\frac{2}{11}$ . Soon , we have : $2$ - $\frac{2}{11}$ = $\frac{20}{11}$ , dividing the numerator by $n$ , we have: $\boxed{\frac{10}{11}}$ . For $B$ we have: $\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + ... + $\frac{1}{7}$ - $\frac{1}{8}$ . So, $1$ - $\frac{1}{8}$ = $\boxed{\frac{7}{8}}$ . Applying $33A - 32B$ = $33*\frac{10}{11}$ - $32*\frac{7}{8}$ = $30 - 28$ = $\boxed{2}$