An algebra problem by Yong Hao Tham

The denominators of this series are of the form $\dfrac{n(n + 1)}{2}$ from $n = 2$ to $n = 24$ . The desired sum is then

$\displaystyle\sum_{n=2}^{24} \dfrac{2}{n(n + 1)} = 2\sum_{n=2}^{24} \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right) = 2\left(\dfrac{1}{2} - \dfrac{1}{25}\right) = 2*\dfrac{(25 - 2)}{50} = \boxed{\dfrac{23}{25}}$ ,

where the last sum telescopes, leaving us with just the first element of the first term and the second element of the last term.