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 $x_{2014}$
I am constantly failing to eliminate $n$ to get $x_{3n+1}$ in the desired form.
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From the two ways to break down $x_{6n-1}$, I find $x_{2n}=4x_n-2n$ – Empy2 Nov 13 '15 at 12:00
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@Michael Don't you mean $x_{6n+1}$? – rubik Nov 13 '15 at 12:09
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$6n-1=3(2n-1)+2=2(3n-1)+1$, and $2n-1=2(n-1)+1,3n-1=3(n-1)+2$ – Empy2 Nov 13 '15 at 12:11
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@Michael Ahh thanks! – rubik Nov 13 '15 at 12:32
1 Answers
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$x_{2n+1}=4x_n+2n+2$, $x_{3n+2}=3x_{n+1}+6x_n$
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....
Martin Sleziak
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Empy2
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1Michael, note that, for a day, until about an hour before you answered, the first given equation read $x_{2n}= 4 x_n + 2n + 2.$ Furthermore, the change was made in a comment to mathlove after he posted an answer showing the problem statement was simply wrong, and mathlove edited the original question....Ah: the OP did the question body, mathlove did the title. – Will Jagy Nov 13 '15 at 17:09
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1http://math.stackexchange.com/revisions/b028b4eb-9d0c-41e3-a0f7-0dbf4d0af31e/view-source so the $x_{2n}$ was there from the start. – Will Jagy Nov 13 '15 at 17:16
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