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Having
$a_n = 3a_{n-1} + 2a_{n-2} + 3·2^{2n-1}$
$a_1 = 12$
$a_0 = 0$

I solved the homogeneous part and got:
$a^{{h}}_n = 1/12·2^n - 1/12·1^n$

This is the particular solution that I need to solve:
$a^{p}_n = C·3·2^{2n-1}$
but the exponent 2n-1 confuses me. I'm don't know how to solve it.

I appreciate any help

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  • $\begingroup$ Note that $2^2=4$ $\endgroup$ Commented Sep 21, 2014 at 20:16
  • $\begingroup$ @MarkBennet If I change it to $4^{n-1}$ and use 2 instead of n, I don't get the same results as with $2^{2n-1}$. Can you elaborate? Thanks $\endgroup$
    – luisc29
    Commented Sep 21, 2014 at 20:37
  • $\begingroup$ $2^{2n-1}=2^{-1}\cdot 2^{2n}=\frac 12 4^n$ - the key thing is to get the exponent $n$ and absorb the other factors into the coefficient. $\endgroup$ Commented Sep 21, 2014 at 20:45

1 Answer 1

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We try for a particular solution of the shape $c\cdot 2^{2n-1}$. Substituting in our recurrence, we have $$c\cdot 2^{2n-1}=3c\cdot 2^{2n-3}+2c\cdot 2^{2n-5}+3\cdot 2^{2n-1}.$$ Dividing by $2^{2n-5}$ we get $$16c=12c+2c+48.$$

Remark: The algebra may feel simpler if we look for a solution of the shape $d\cdot 2^{2n}$, that is, $d\cdot 4^n$. We get $d\cdot 4^{n}=3d\cdot 4^{n-1}+2d\cdot 4^{n-2}+\frac{3}{2}\cdot 4^n$. Divide through by $4^{n-2}$.

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  • $\begingroup$ Thanks, this is very clear. But the solution shouldn't be $c·3·2^{2n-1}$? Taking into consideration that the non-homogeneous part is $3·2^{2n-1}$. Thanks $\endgroup$
    – luisc29
    Commented Sep 21, 2014 at 20:43
  • $\begingroup$ The constant term is plain $3\cdot 2^{2n-1}$, so that is what I have in the equation. Solve for $c$. We get $c=24$. So a particular solution is $24\cdot 2^{2n-1}$, or if you prefer $12\cdot 2^{2n}$. $\endgroup$ Commented Sep 21, 2014 at 21:55

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