solution
$1+2+3+4+5+6........n=\frac{n(n+1)}{2}$
$1 + \frac {1}{(1+2)} + \frac {1}{(1+2+3)}+...+ \frac {1}{(1+2+3+4+...+20)}$
$2 \cdot (\frac {1}{(1)(2)} + \frac {1}{(2)(3)} + \frac {1}{(3)(4)}+...+ \frac {1}{(20)(21)})$
$2 \cdot (\frac {1}{1}-\frac {1}{21})$
$\frac {40}{21}$
$1\frac{19}{21} $
