Domain and range transformation

Question

How can I solve this recurrence relation using Domain and Range transformations:

$$ \begin{array}{rcl} n^2 a_n &=& 5(n-1)^2 a_{n-1} +2 \\ a_0 &=& 0 \\ \end{array} $$

Answer 1

Let $n^2a_n = x_n$. Then the given relation can be rewritten as $x_n = 5x_{n-1}+2$ or equivalently, $x_{n+1} -6x_{n}+5x_{n-1} = 0$. Then $x_n = c_15^n+c_2$ for constants $c_1, c_2$ determined by the initial conditions.

Then it is easy to get a closed form for $a_n$.

Answer 2

Use generating functions on the recurrence given by Sandeep Silval: Define $g(z) = \sum_{n \ge 0} x_n z^n$, write the recurrence as: $$ x_{n + 1} = 5 x_n + 2 $$ Multiply by $z^n$, sum over $n \ge 0$, and recognize the resulting sums: $$ \frac{g(z) - x_0}{z} = 5 g(z) + \frac{2}{1 - z} $$ As partial fractions: $$ g(z) = \frac{1}{2 (1 - 5 z)} - \frac{1}{2 (1 - z)} $$ thus: \begin{align} x_n &= \frac{5^n - 1}{2} \\ a_n &= \begin{cases} 0 & n = 0 \\ \displaystyle \frac{5^n - 1}{2 n^2} & n > 0 \end{cases} \end{align} \end{align}