Help in solving linear recurrence relation I need to solve the following recurrence relation: $a_{n+2} + 2a_{n+1} + a_n = 1 + n$

My solution:
Associated homogeneous recurrence relation is:
$a_{n+2} + 2a_{n+1} + a_n = 0$
Characteristic equation: $r^2 + 2r + 1 = 0$
Solving the characteristic equation, we get: $r = -1$ with multiplicity $m = 2$
Therefore, solution of the homogeneous recurrence relation is:
$a_n^{(h)} = (c_1 + c_2n)(-1)^n$ 
Let the particular solution of the given equation be
$a_n = c_3 + c_4n$
since, $(n + 1)$ is polynomial of degree 1.
Substituting in the given equation, we get:
$c_3 + c_4(n + 2) + 2(c_3 + c_4(n + 1)) + c_3 + c_4n = n + 1$
Comparing the corresponding coefficients, we get: $c_4 = 1/4$ and $c_3 = 0$.
Therefore, $a_n^{(p)} = n / 4$
Hence, the solution, would be:
$a_n = (c_1 + c_2n)(-1)^n + n / 4$
But the solution in textbook is 
$a_n = (c_1 + c_2n)(-1)^n + 1/6(2n - 1)$
Please explain me where I am going wrong.
Thanks!

General technique uses generating functions. Define:
$\begin{equation*}
  A(z) = \sum_{n \ge 0} a_n z^n
\end{equation*}$
Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$\begin{align*}
  \sum_{n \ge 0} a_{n + 2} z^n
    + 2 \sum_{n \ge 0} a_{n + 1} z^n
    + \sum_{n \ge 0} a_n z^n
    &= \sum_{n \ge 1} z^n
         + \sum_{n \ge 0} n z^n \\
  \frac{A(z) - a_0 - a_1 z}{z^2}
    + 2 \frac{A(z) - a_0}{z}
    + A(z)
    &= \frac{1}{1 - z}
         + z \frac{d}{d z} \frac{1}{1 - z} \\
    &= \frac{1}{1 - z} + \frac{z}{(1 - z)^2}
\end{align*}$
Solve for $A(z)$, split into partial fractions:
$\begin{equation*}
  A(z)
    = -\frac{4 a_1 + 4 a_0 - 1}{(1 + z)^2}
        + \frac{\frac{4 a_1 + 8 a_0 - 1}{(1 + z)}
        + \frac{1}{4 (1 - z)^2}
        - \frac{1}{4 (1 - z)}
\end{equation*}$
Note that:
$\begin{align*}
  (1 - \alpha z)^{-m}
    &= \sum_{n \ge 0}
         (-1)^n \binom{-m}{n} \alpha^n \zeta^n \\
    &= \sum_{n \ge 0}
         \binom{m + n - 1}{m - 1} \alpha^n z^n
\end{align*}$
Also:
$\begin{align*}
  \binom{n + 1}{1}
    &= n + 1 \\
  \binom{n + 0}{0}
    &= 1
\end{align*}$
and you can read off the coefficients. The solution here has the form:
$\begin{equation*}
  a_n
    = (-1)^n c_1 n + (-1)^n c_2 + c_3 n + c_4
\end{equation*}$
