Find $x_{2014}$ in the recurrence $x_{2n+1}=4x_n+2n+2$ and $x_{3n+2}=3x_{n+1} + 6x_{n}$

Question

Given: $x_{2n+1}=4x_n+2n+2$ and $x_{3n+2}=3x_{n+1} + 6x_{n}$ for all $n\in\mathbb{N}$.

Prove that: $x_{3n+1}=x_{n+2}-2x_{n+1}+10x_{n}$ and hence find $x_{2014}$

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I am constantly failing to eliminate $n$ to get $x_{3n+1}$ in the desired form.

Answer 1

$x_{2n+1}=4x_n+2n+2$, $x_{3n+2}=3x_{n+1}+6x_n$
From $x_{6n-1}=3x_{2n}+6x_{2n-1}=3x_{2n}+24x_{n-1}+12n$ and $x_{6n-1}=4x_{3n-1}+6n=12x_{n}+24x_{n-1}+6n$, I find $x_{2n}=4x_{n}-2n$.

Use the rules for $x_{2n}$and $x_{2n+1}$ to build everything in terms of x(1), then use the rule for $x_{3n+2}$ to check.

Let $x_{1}=a x_{2}=4a-2,x_{3}=4a+4,x_{4}=16a-12,x_{5}=16a-2$ should equal 12a-6+6a, so $a=2$.

$x_{i}=2,6,12,20,30$, which is enough for a guess....