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!
 A: I have verified your results using a slightly different method. Here I reduce the original recurrence to a more familiar one with a known solution. Thus consider
$$a_{n+2} + 2a_{n+1} + a_{n} = 1+n$$
Let $a_{n}=f_n+pn+q$, so that
$$
\begin{align}
&f_{n+2}+p(n+2)+q+\\
&2f_{n+1}+2p(n+1)+2q+\\
&f_{n}+pn+q=1+n
\end{align}
$$
Now chose $p$ and $q$ so that those terms vanish, to wit,
$$
p=\frac{1}{4}\\
q=0;
$$
So that we are left with
$$f_n=-2f_{n-1}-f_{n-2}$$
with characteristic roots $(-1,-1)$ and a solution given by
$$f_n=\left(nf_1+(n-1)f_0\right)(-1)^{n-1}$$
where $f_0=a_0$ and $f_1=a_1-p$. The complete solution is then given by
$$a_n=f_n+pn$$
I have verified this solution for arbitrary values of $a_0$ and $a_1$.
A: 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:
$\begin{equation*}
  a_n
    = (-1)^n c_1 n + (-1)^n c_2 + c_3 n + c_4
\end{equation*}$
