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$f(n)=6f(n-1)-9f(n-2)+(n^2+1)3^n$

The root for the above relation is 3 two times. So its general term will be:

$f(n) = c_{1}3^n + c_{2}n3^n + something$

According to my notes $something: n^2(n^2p_{2,0}+np_{1,0}+p_{0,0})3^n = (n^2+1)3^n$

That should have been correct if it was: $+n^23^n$
But it is: $+(n^2+1)3^n = 3^nn^2+3^n$
so it should be:   $n^2(n^2p_{2,0}+np_{1,0}+p_{0,0})3^n + n^2(p_{0,1})3^n=(n^2+1)3^n$

Is my reasoning wrong?

Complex1238877
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  • Following http://math.stackexchange.com/questions/807306/solving-a-recursion-relation-a-n1-3a-n4a-n-12/807343#807343 Let $f(m)=3^mg(m)$

    $$\implies g(n)=2g(n-1)-g(n-2)+n^2+1$$

    Let $g(r)=h(r)+a_0+a_1r+a_2r^2+\cdots$

    $$\implies h(n)+a_0+a_1n+a_2n^2+a_3n^3+\cdots$$ $$=2[h(n-1)+a_0+a_1(n-1)+a_2(n-1)^2+a_3(n-1)^3+\cdots]-[h(n-2)+a_0+a_1(n-2)+a_2(n-2)^2+a_3(n-2)^3+\cdots]+n^2+1$$

    $$=2h(n-1)-h(n-2)+[n^3(a_3)+n^2(a_2+1)]$$ Now compare the coefficients of $n,n^2.n^3,\cdots$

     – lab bhattacharjee May 02 '15 at 13:44

1 Answers1

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i will try a change of variable $$u_n = f_n3^{-n}, \quad f_n = 3^nu_n.$$ the recurrence relation for $u_n$ is $$3^nu_n = 6\, 3^{n-1}u_{n-1} - 9\, 3^{n-2}u_{n-2}+(n^2+1)\, 3^n $$ dividing out by $3^n,$ gives you $$Lu_n = u_n- 2u_{n-1}+u_{n-2}= n^2+1 \tag 1$$

we see that $$\begin{align}L(n+1)^4 &= (n+1)^4 -2n^4 +(n-1)^4=12n^2+2,\\ L(n+1)^2 &=(n+1)^2-2n^2+(n-1)^2= 2\end{align}$$ therefore a particular solution is $$u_n = \frac1{12}\left((n+1)^4 -1\right), \quad f(n) = \frac1{12}\left((n+1)^4 -1\right)3^n.$$

abel
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