Finding the explicit formula for a recursive sequence $a$ where $a_0=3$, $a_1=6$, $a_2=14$, and $a_{n+3}= 6a_{n+2} - 11 a_{n+1}+6a_n $ for $n = 0,1,2\ldots$ $$ S(x)= \sum_{n=0}^\infty a_n x^n $$
Prove that the radius of convergence is $R>0$ and $S(x)(1-x)(1-2x)(1-3x), x \in \mathbb{R}$ is a square trinomial (find coefficients).
Please a hint, how to do it?
You have what is known as a linear homogeneous recurrence relation with constant coefficients, also saddled with the acromym LHRRWCC. Achille has given you the form of the solution to these critters. Continuing from that comment, once you've found $\lambda_1, \lambda_2, \lambda_3$ and $\alpha_1, \alpha_2, \alpha_3$ (hint: all six values are really simple), use these to find $S(x)$ by writing $$\begin{align} S(x) &=\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty (\alpha_1\lambda_1^n+\alpha_2\lambda_2^n+\alpha_3\lambda_3^n)x^n\\ &=\sum_{n=0}^\infty\alpha_1(\lambda_1x)^n+\sum_{n=0}^\infty\alpha_2(\lambda_2x)^n+\sum_{n=0}^\infty\alpha_3(\lambda_3x)^n &\text{(Why?)}\\ &=\frac{\alpha_1}{1-\lambda_1x}+\frac{\alpha_2}{1-\lambda_2x}+\frac{\alpha_3}{1-\lambda_3x} &\text{(Why?)} \end{align}$$ Knowing what $a_n$ is, the problem expects you to know how to compute the radius of convergence and show that it is greater than zero. This isn't too hard.
Also, knowing what $S(x)$ is, a little algebra should show you that $$ S(x)(1-x)(1-2x)(1-3x) $$ has the form $px^2+qx+r$. As a check for your work, you should discover that $$ S(x)(1-x)(1-2x)(1-3x)=11x^2-12x+3 $$ Hope this helps. If not, feel free to ask for clarification.
hint: $$\lambda^3-6\lambda^2+11\lambda-6=\lambda^3-(3+2+1)\lambda^2+(3\cdot2+3\cdot1+2\cdot1)\lambda-3\cdot2\cdot1$$
