A complicated series

$S$ can be represented as $\frac{1+2}{1 \times 2 \times 3} + \frac{2+3}{2 \times 3 \times 4} +......\frac{18+19}{18 \times 19 \times 20}$ which can be simplified to $\frac{1}{2 \times 3} + \frac{1}{1 \times 3} + \frac{1}{3 \times 4} + \frac{1}{2 \times 4 } ......... \frac{1}{19 \times 20} + \frac{1}{18 \times 20}$ This can be further simplified into three different series $\frac{1}{2 \times 3} + \frac{1}{3 \times 4} + ......... + \frac{1}{19 \times 20} ------ (1)$ $\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + ...... \frac{1}{17 \times 19} ------(2)$ $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + ......... \frac{1}{18 \times 20} ------(3)$ Then,we can calculate the sum of these three series.Series (1) is equal to $\frac{1}{2} - \frac{1}{20} = \frac{9}{20}$ .Series (2) is equal to $\frac{1}{2} \times (\frac{1}{1}-\frac{1}{19}) = \frac{9}{19}$ .Series (3) is equal to $\frac{1}{2} \times(\frac{1}{2} - \frac{1}{20}) = \frac{9}{40}$ .Adding these up we get $S = \frac{873}{760} = 1\frac{113}{760}$ .So $a + b + c = 1 + 113 + 760 = 874$ .

We have $S = \sum^{18}_{n=1} \frac{2n+1}{n(n+1)(n+2)} = \sum^{18}_{n=1} \frac{1}{n+2} \left( \frac{1}{n} + \frac{1}{ n+1} \right)$ $= \sum^{18}_{n=1} \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right)+ \left(\frac{1}{n+1} - \frac{1}{n+2} \right)$ $= \frac{1}{2} \sum^{18}_{n=1} \left( \frac{1}{n} - \frac{1}{n+1} \right)+ \frac{3}{2} \sum^{18}_{n=1} \left(\frac{1}{n+1} - \frac{1}{n+2} \right)$ $= \frac{1}{2} \left( 1 - \frac{1}{19} \right)+ \frac{3}{2} \left( \frac{1}{2} - \frac{1}{20} \right) = 1 \frac{113}{760}$ Thus, we have $a=1, b = 113, c = 760 \Rightarrow a + b + c = \boxed{874}$

rth term tr = (2r+1)/r(r+1)(r+2) where r = 1 to 18 solving we get S = 873/760 or 1/113/760 so a + b + c =874