Question

(4+8+12+16+...+204):(2+4+6+...+102=

Answer 1

Both numerator and denominator are a sum of 51 terms, counted skipping 4s (numerator) and 2s (numerator).

So, you have

$\dfrac{\displaystyle\sum_{i=1}^{51} 4i}{\displaystyle\sum_{i=1}^{51} 2i}$

Use the linearity of sum to rewrite this as

$\dfrac{\displaystyle4\sum_{i=1}^{51} i}{\displaystyle2\sum_{i=1}^{51} i}$

So, as you can see, the numerator is exactly twice the numerator: you can cancel out the sum and you have

$\dfrac{4}{2} = 2$