Using generation functions solve the following difference equation

Question

Using generation functions solve the following difference equation

$$ a_{n+1} - 3a_{n+2} + 2a_n = 7n ; n\geq0; a_0 = -1; a_1 = 3. $$

Answer 1

Thanks for the hint was able to solve it. This was my solution.

$$ a_{n+1} - 3a_{n+2} + 2a_n = 7n ; n\geq0; a_0 = -1; a_1 = 3. $$

$$ (\frac{a{(z)} - a_{0} + za_1}{z^2}) - 3(\frac{a{(z)} - a_{0}}{z^2}) + 2a(z) = \frac{7z}{(1 - z)^2}; a_0 = -1; a_1 = 3. $$

Substitute values in:

$$ (\frac{a{(z)} + 1 - 3z}{z^2}) - 3(\frac{a{(z)} + 1}{z^2}) + 2a(z) = \frac{7z}{(1 - z)^2}; a_0 = -1; a_1 = 3. $$

Expand and multiply through by $z^2$

$$ a(z) + 1 -3z - 3a(z)z -3z + 2a(z)z^2 = \frac{7z^3}{(1 - z)^2} $$

$$ a(z)[1 -3z + 2z^2] = \frac{7z^3}{(1 - z)^2} + 6z - 1 $$

$$ a(z) = \frac{13z^3 - 13z^2 + 8z -1}{(z-1)^2(2z-1)}$$

By Partial Fraction decomposition:

$$ a(z) = \frac{-11}{(2z-1)} + \frac{12}{(z-1)} + \frac{7}{(z-1)^2} + \frac{7}{(z-1)^3}$$

By Extracting coefficients we get: $$ a(z) = \frac{-7[n(n+1)]}{2} + 11 * 2^n - 12$$

Answer 2

Define the generating function $A(z) = \sum_{n \ge 0} a_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$, Recognize, e.g.: \begin{align} \sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2} \\ \sum_{n \ge 0} z^n &= \frac{1}{1 - z} \\ \sum_{n \ge 0} n z^n &= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \end{align} Solve the result for $A(z)$, write as partial fractions (a computer algebra system helps), and read off the coefficients by using geometric series or: $$ (1 + u)^{-m} = \sum_{k \ge 0} \binom{-m}{k} u^k = \sum_{k \ge 0} \binom{k + m - 1}{m - 1} (-1)^k u^k $$