Both numerator and denominator are a sum of 51 terms, counted skipping 4s (numerator) and 2s (numerator).
So, you have
[tex] \dfrac{\displaystyle\sum_{i=1}^{51} 4i}{\displaystyle\sum_{i=1}^{51} 2i} [/tex]
Use the linearity of sum to rewrite this as
[tex] \dfrac{\displaystyle4\sum_{i=1}^{51} i}{\displaystyle2\sum_{i=1}^{51} i} [/tex]
So, as you can see, the numerator is exactly twice the numerator: you can cancel out the sum and you have
[tex] \dfrac{4}{2} = 2 [/tex]
