How to find polynomials $a(x)$ and $b(x)$ such that $c(x) = a(x) / b(x)$?

Question

Consider the sequence $c_0, c_1, c_2,\ldots$ satisfying $c_i =2\cdot 3^i − i^2\cdot(−1)^i$.
Let $c(x) = c_0 + c_1x + c_2x^2 + \ldots$
Find polynomials $a(x)$ and $b(x)$ such that $c(x) = a(x) / b(x)$.

Can someone tell me where should I start this question or how should I solve this question please?
Thanks in advance.

Answer 1

Don't worry so much about the $\frac{a(x)}{b(x)}$ part -- the main difficulty of the problem is finding a form for that ugly looking polynomial.

Let $d_i = 2*3^i$ and $e_i = (-1)^i i^2$. Then $c_i = d_i - e_i$. Define $d(x)$ and $e(x)$ similarly.

$d(x) = 2 + 6x + 18x^2 + 54x^3 + \dots = \frac{2}{1 - 3x}$

Next, $\displaystyle \sum_{i = 0}^{\infty} (-x)^i = \frac{1}{1+x}$. Differentiate to get $\displaystyle \sum_{i = 0}^{\infty} -i(-x)^{i-1} = \frac{-1}{(x+1)^2}$. Multiply by $-x$ and differentiate to get $\displaystyle \sum_{i = 0}^{\infty} i^2(-x)^{i-1} = \frac{x-1}{(x+1)^3}$. Multiply once more by $-x$ to get $$e(x) = \frac{x(x-1)}{(x+1)^3}$$

So, $c(x) = d(x) - e(x) = \large{-\frac{5x^2 + 2x^2 + 7x + 2}{(x+1)^3(3x-1)}}$

This holds for $|x| < 1$

If you want that holds for all $x$... how about $\frac{\displaystyle \sum_{i=0}^{\infty} (2 * 3^i - i^2 * (-1)^i)x^{i+1}}{x}$?

Answer 2

Hint. You can add up a geometric series: $$\frac{1}{1-x}=\sum_{i=0}^\infty x^i\ .$$ Differentiating both sides, $$\frac{1}{(1-x)^2}=\sum_{i=0}^\infty ix^{i-1}\ .$$ Multiply by $x$ and differentiate again: $$\frac{x}{(1-x)^2}=\sum_{i=0}^\infty ix^i\ ,\quad\hbox{so}\quad \frac{1+x}{(1-x)^3}=\sum_{i=0}^\infty i^2x^{i-1}\ .$$ Now you want to evaluate $$\sum_{i=0}^\infty (2\times3^i-i^2(-1)^i)x^i\ .$$ See if you can do this using the above results.