Consider a sequence of positive integers $x_n$ such that: \[(\text{A})\ x_{2n+1}=4x_n+2n+2 \] \[(\text{B})\ x_{3n+\color[rgb]{0.9529,0.0980,0.0118}2}=3x_{n+1}+6x_n \] for all $n\ge 0$. Prove that \[(\text{C})\ x_{3n-1}=x_{n+2}-2x_{n+1}+10x_n \] for all $n\ge 0$.
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Tags: algebra proposed, algebra
spanferkel
31.10.2010 00:50
WakeUp wrote: Consider a sequence of positive integers $x_n$ such that: \[(\text{A})\ x_{2n+1}=4x_n+2n+2 \] \[(\text{B})\ x_{3n+3}=3x_{n+1}+6x_n \] for all $n\ge 0$. Prove that \[(\text{C})\ x_{3n-1}=x_{n+2}-2x_{n+1}+10x_n \] for all $n\ge 0$. Is there a typo or am I wrong? $x_1=4x_0+2$ $x_3=4x_1+4$ $x_3=3x_1+6x_0$ This system yields $x_0=3,x_1=14,x_3=60$, further $x_7=248$ and $x_{15}=1008$. $x_5=4x_2+6$ $1008=3x_5+6x_4$ $x_9=4x_4+10=3x_3+6x_2$, thus $4x_4=170+6x_2$ This system yields $x_2=35,x_4=95,x_5=146$. But that doesn't fit with (C) for $n=1$
WakeUp
31.10.2010 02:14
I apologise, I checked it at least 4 times prior to posting, and I still made a mistake. Sorry for wasting your time. I've edited the problem.
sarti25
10.02.2015 20:10
So has anyone solved this ? Please help me ...
Yor
18.07.2017 11:12
By putting n=0 in (A), we get x1=4*x0+2...................(1) By putting n=0 in (B), we get x2=3*x1+6*x0..............(2) By putting n=2 in (A), we get x5=4*x2+6...................(3) By putting n=1 in (B), we get x5=3*x2+6*x1..............(4) By putting n=1 in (A), we get x3=4*x1+4..............(2.2) (3)=(4), so 4*x2+6=3*x2+6*x1 => x2=6*x1-6...........(5) (5)=(2), so 6*x1-6=3*x1+6*x0 => 3*x1=6*x0+6 => x1=2*x0+2.........(6) (6)=(1), so 2*x0+2=4*x0+2 => 2*x0=4*x0 => x0=0, putting in (1),(2),(2.2) we get x1=2, x2=6, x3=12 Putting n=1 in (C), we get x2=x3-2*x2+10*x1 , inserting numbers we get 6=12-2*6+12*2 => 6=24 contradiction. Something is missing in (C).
Yor
18.07.2017 12:23
Lets find out what is missing in (C), and let the missing part be 'A'. So, \[(\text{C})\ x_{3n-1}=x_{n+2}-2x_{n+1}+10x_n +A\]putting n=>(3n+2) in (A), we get \[x_{6n+5}=4x_{3n+2}+6n+6...............(1)\]and n=>(2n+1) in (B), we get \[x_{6n+5}=3x_{2n+2}+6x_{2n+1}...............(2)\](1)=(2), so \[3x_{2n+2}+6x_{2n+1}=4x_{3n+2}+6n+6\]using (A) and (B), we get \[3x_{2n+2}+6x_{2n+1}=4x_{3n+2}+6n+6\]\[3x_{2n+2}+6(4x_n+2n+2)=4(3x_{n+1}+6x_n )+6n+6\]simplifying, \[4x_{n+1}=x_{2n+2}+2n+2..............(3)\]put n=>n+1 in (A) \[x_{2n+3}=4x_{n+1}+2n+4\]=> \[4x_{n+1}=x_{2n+3}-2n-4\]put in (3) \[x_{2n+2}+2n+2=x_{2n+3}-2n-4\]\[x_{2n+3}=x_{2n+2}+4n+6\]put n=>(n-3)/2 \[x_{n}=x_{n-1}+2n.............(4)\], and from (4) we get: \[x_{n-1}=x_{n}-2n\]and \[x_{n+2}=x_{n+1}+2n+4\]and \[x_{n+1}=x_{n}+2n+2\]. And putting n=>n-1 in (B), we get \[x_{3n-1}=3x_{n}+6x_{n-1}\]so (C) will be \[ x_{3n-1}=x_{n+2}-2x_{n+1}+10x_n +A\]\[ (3x_{n}+6x_{n-1})=(x_{n+1}+2n+4)-2x_{n+1}+10x_n +A\]\[ 3x_{n}+6(x_{n}-2n)=2n+4-x_{n+1}+10x_n +A\]\[ 9x_{n}-12n=2n+4-(x_{n}+2n+2)+10x_n +A\]\[ 9x_{n}-12n=2n+4-x_{n}-2n-2+10x_n +A\]\[ -12n-2=-x_{n}+10x_n-9x_{n} +A\]\[ A=-12n-2\]Therefore (C) shold be \[ x_{3n-1}=x_{n+2}-2x_{n+1}+10x_n -12n-2\]
huashiliao2020
08.05.2023 04:46
Yor wrote:
Lets find out what is missing in (C), and let the missing part be 'A'. So, \[(\text{C})\ x_{3n-1}=x_{n+2}-2x_{n+1}+10x_n +A\]putting n=>(3n+2) in (A), we get \[x_{6n+5}=4x_{3n+2}+6n+6...............(1)\]and n=>(2n+1) in (B), we get \[x_{6n+5}=3x_{2n+2}+6x_{2n+1}...............(2)\](1)=(2), so \[3x_{2n+2}+6x_{2n+1}=4x_{3n+2}+6n+6\]using (A) and (B), we get \[3x_{2n+2}+6x_{2n+1}=4x_{3n+2}+6n+6\]\[3x_{2n+2}+6(4x_n+2n+2)=4(3x_{n+1}+6x_n )+6n+6\]simplifying, \[4x_{n+1}=x_{2n+2}+2n+2..............(3)\]put n=>n+1 in (A) \[x_{2n+3}=4x_{n+1}+2n+4\]=> \[4x_{n+1}=x_{2n+3}-2n-4\]put in (3)
\[x_{2n+2}+2n+2=x_{2n+3}-2n-4\]\[x_{2n+3}=x_{2n+2}+4n+6\]put n=>(n-3)/2 \[x_{n}=x_{n-1}+2n.............(4)\], and from (4) we get:
\[x_{n-1}=x_{n}-2n\]and \[x_{n+2}=x_{n+1}+2n+4\]and \[x_{n+1}=x_{n}+2n+2\].
And putting n=>n-1 in (B), we get \[x_{3n-1}=3x_{n}+6x_{n-1}\]so (C) will be
\[ x_{3n-1}=x_{n+2}-2x_{n+1}+10x_n +A\]\[ (3x_{n}+6x_{n-1})=(x_{n+1}+2n+4)-2x_{n+1}+10x_n +A\]\[ 3x_{n}+6(x_{n}-2n)=2n+4-x_{n+1}+10x_n +A\]\[ 9x_{n}-12n=2n+4-(x_{n}+2n+2)+10x_n +A\]\[ 9x_{n}-12n=2n+4-x_{n}-2n-2+10x_n +A\]\[ -12n-2=-x_{n}+10x_n-9x_{n} +A\]\[ A=-12n-2\]Therefore (C) shold be \[ x_{3n-1}=x_{n+2}-2x_{n+1}+10x_n -12n-2\]