Consider the following expression :

11+2+3+..+n=1n(n+1)2=2n(n+1)

So we have :

11+2+11+2+3+11+2+3+4+...+11+2+3+4+...+50

=22.3+23.4+24.5+...+250.51

=2(12.3+13.4+14.5+...+150.51)

=2(12−13+13−14+14−15+...+150−151)

=2(12−151)=1−251=4951

Consider the following expression :

\(\dfrac{1}{1+2+3+..+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}\)

So we have :

\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+50}\)\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{50.51}\)

\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{50.51}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{51}\right)=1-\dfrac{2}{51}=\dfrac{49}{51}\)