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I have the following recurrent equation

$ a_1=3\\a_2=1\\a_n=a_{n-1}+2a_{n-2}+2 $

and I cannot seem to be able to solve it correctly. According to wolframalpha and my math textbook (that contains results but not the actual process), the result should be $a_n=-2(-1)^n+2^n-1$, but I cannot seem to actually get that result and I cannot find where I am making a mistake.

My current approach (based on the textbook) is:

$a_{n=1} = 2$, needs to add 1.

$a_{n=2} = 5$, needs to substract 4.

The universal equation therefore is $a_n=a_{n-1}+2a_{n-2}+2+[n=1]-4[n=2]$.

$A(x)=xA(x) + 2x^2A(x)+2+x-4x^2$, divide both sides by $A(x)$:

$1= x + 2x^2 + {\frac {2+x-4x^2} {A(x)}}$, move $A(x)$ to one side:

$A(x) = {\frac {-4x^2+x+2} {-2x^2 -x +1}}$, divide the polynomials:

$A(x) = 2+{\frac {3x} {-2x^2-x+1}}$, split into multiple fractions:

$A(x) = 2+3x * \frac {1} {1-2x} * \frac {1} {1+x}$.

This is the point where I get lost, how exactly am I supposed to expand something like this into a power series? If I made a mistake in the process, could someone explain, why? Does anyone have other solutions for this problem? Thanks a lot.

nonuser
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kubci98
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1 Answers1

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Well I would do like this. Since for all $n$ we have: $$ a_n-a_{n-1}-2a_{n-2}=2 $$ we have also (change $n$ to $n+1$) $$a_{n+1}-a_{n}-2a_{n-1}=2$$ so if we substract these two we get: $$a_{n+1}-2a_n-a_{n-1}+2a_{n-2}=0$$ so zeroes of characteristic equation $x^3-2x^2-x+2=0$ are $1,2,-1$ so $$a_n = a\cdot (-1)^n+b\cdot 2^n+c\cdot 1^n$$

nonuser
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    so $a$,$b$,$c$ are just constants, that I would then have to calculate myself, right (which means solving a simple multiple equations system)? Thanks, this seems much easier than my approach. – kubci98 May 27 '18 at 15:31
  • Yes............ – nonuser May 27 '18 at 15:40
  • OK, I am probably really just stupid at this point, but I tried doing this and it did not yield the desired result:

    $a_1=3$, $a_2=1$, $a_3=3+2*2+2 = 7$, so $3=-a+2b+c$, $1=a+4b+c$, $7=-a+8b+c\$.

    The results should be a=-2; b=1 and c=-1, but the system above returns a=-5/3, b=2/3 and c=0.

     – kubci98 May 27 '18 at 15:53
  • $a_3=9$ not 7... – nonuser May 27 '18 at 15:55
  • oh my. yep, my mistake, sorry. that gives the right result. – kubci98 May 27 '18 at 15:56