Q:

Continuable fractions

Let $X$ and $Y$ be positive integers such that the first few digits of the fractions
$\dfrac{29}{{\color{red}{70}}},\ \dfrac{{\color{red}{70}}}{{\color{blue}{169}}},\ \dfrac{{\color{blue}{169}}}{{\color{green}{408}}},\ \dfrac{{\color{green}{408}}}{{\color{brown}{X}}},\ \dfrac {{\color{brown}{X}}}{{\color{grey}{Y}}} $
are all equal to $0.4142$. What is ${{\color{brown}{X}}}+{{\color{grey}{Y}}}?$
Note: The fractions are not equal.

A:

A useful trick for approximating the fractional part of $\sqrt{2}$ is the following sequence
$\frac{p}{q}\implies \frac{q}{p+2q}$
Clearly, the sequence in question follows this sequence as shown below
$\frac{29}{70} \implies \frac{70}{29 + (2)(70)} = \frac{70}{169} \implies \frac{169}{70 + (2)(169)} = \frac{169}{408} $
Therefore, the sequence containing $X$ and $Y$ is as follows
$\frac{169}{408}, \frac{408}{985}, \frac{985}{2378}$
Thus, $X + Y = 985 + 2378 = \boxed{3363}$

