Find the value of $\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90}$ Find the value of $p+q$, where $p$ and $q$ are two positive integers such that $p$ and $q$ have no common factor larger than $1$ and 
$$\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} = \frac{p}{q}.$$
By using Wolfram Alpha, I obtain the sum is $\frac{3}{20}$. But I have no idea on how to obtain the sum.
Any hint would be appreaciated. 
[The question is taken from SMO 2015 Junior section]
 A: We can observe that
\begin{align}
20&=4\cdot 5\\
30&=5\cdot6\\
42&=6\cdot7\\
56&=7\cdot8\\
72&=8\cdot9\\
90&=9\cdot10.
\end{align}
Can you use partial fraction?
A: We  know that 
$$
S_n=\frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdots+ \frac{1}{(n-1)n}=1-\frac{1}{n}
$$
Then yours sum is exactly $$S_{10}-S_4=(1-\frac{1}{10})-(1-\frac{1}{4})=\frac{1}{4}-\frac{1}{10}=\frac{5-2}{20}=\frac{3}{20}.$$
A: $1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 = $
$1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9 + 1/9*10 = $
Note:  $\frac{1}{n(n+1)} + \frac{1}{(n +1)(n +2)} = \frac{n+2 + n}{n(n+1)(n+2} = \frac {2n + 1}{n(n+1)(n+2)} = \frac{2}{n(n+2)}$
And $\frac{k}{n(n+k)} + \frac{1}{(n+k)(n + k + 1)} = \frac{k(n + k + 1) + n}{n(n+k)(n + k + 1)} = \frac{(k + 1)n + (k+1)k}{n(n+k)(n+k + 1)} = \frac{k+1}{n(n+ k + 1)}$.
So... 
$1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9 + 1/9*10 = $
$2/4*6 + 1/6*7 + 1/7*8 + 1/8*9 + 1/9*10 = $
$3/4*7 + 1/7*8 + 1/8*9 + 1/9*10 = $
$4/4*8 + 1/8*9 + 1/9*10 = $
$5/4*9 + 1/9*10 = $
$6/4*10 =  3/20 = p/q$
$p + q = 3 + 20 = 23$.
