Solving $X + \frac{X}{1+2} + \frac{X}{1+2+3} + \frac{X}{1+2+3+4} +\dots+ \frac{X}{1+2+3+\dots+2017} = 2017$ $$X + \frac{X}{1+2} + \frac{X}{1+2+3} + \frac{X}{1+2+3+4} +\dots+ \frac{X}{1+2+3+\dots+2017} = 2017$$
I’ve noticed that when you substitute around a bit, you get that
$yX$, where $y$ is the amount of $X$, is equal to $2017 \cdot (1+2+\dots+2017)$.
Does this actually help?

We can prove the following via induction. $$ \dfrac1{1} + \dfrac1{1+2} + \dfrac1{1+2+3} + \cdots + \dfrac1{1+2+\cdots+n} = \dfrac {2n}{n+1} $$
So the solution to the equation
$$X + \frac{X}{1+2} + \frac{X}{1+2+3} + \frac{X}{1+2+3+4} +\dots+ \frac{X}{1+2+3+\cdots+n} = n$$ is $$ X = \dfrac{n+1}2 .$$
