Solving non-homogeneous recurrence relations of type $a_{n+1} - a_{n} = c_{_1}n+c_{_2}$ I do not understand how to go about solving the following form of non-homogeneous recurrence relations;
$a_{n+1} - a_{n} = c_{_1}n+c_{_2}$.
I have the following question: 
$a_{n+1} - a_{n} = 2n+3$
The solution to this equation is : $(n+1)^2$
I have tried for quite some time but no method I have attempted leads me to this solution. 
My steps in solving:
$a_n = a_n^h + a_n^p \\ a_n = c(h) + A(p)\\
a_n^h = (h-1)=0 \\
 h=1n$
$a_n^p = A2(n+1) +A2n = 2n +3 \\ 
A = \frac{3}{2} \\
c = -1/2 \\
p = 2n$

$a_n = c(h) + A(p)$ 
$a_n = \frac{-1}{2}n + 3n$
Any help would be much appreciated, and please do not down vote, I have attempted this question for quite some time now and have put a great deal of effort in my attempt to solve it and am here to learn. Thanks all. 
 A: Look for a particular solution of the shape $an^2+bn$. This will be added to the general solution of the homogeneous equation, which in this simple case is just an arbitrary constant $A$.
Substitute. We get 
$$a(n+1)^2+b(n+1)-an^2-bn=c_1n+c_2.$$
Simplify. We get
$$2an+a+b=c_1n+c_2,$$
giving $a=\frac{c_1}{2}$  and $b=c_2-\frac{c_1}{2}$.
For your problem, this gives particular solution $n^2+2n$, not quite the $(n+1)^2$ you mentioned, but just as correct. The general solution is $A+n^2+2n$.
Added: In our case we have $c_1=2$ and $c_2=3$. These are given, since we are solving the recurrence $a_{n+1}-a_n=2n+3$. So from the above equations we have $a=\frac{c_1}{2}=1$ and $b=c_2-\frac{c_1}{2}=3-1=2$.  
Thus the general solution of the inhomogeneous recurrence is $a_n=A +n^2+2n$. To evaluate $A$, we would need an initial condition. If that condition is $a_0=1$, then $A=1$ and the solution is $a_n=n^2+2n+1$, or equivalently $(n+1)^2$. Note that $(n+1)^2$ is also a particular solution of the inhomogeneous equation, since if $f(n)$ is a solution so is $f(n)+k$.
A: $a_2-a_1=5$
$a_3-a_2=7$
...
$a_{n} - a_{n-1} = 2n+1$
Summing it up gives $a_n-a_1=5+7+...+(2n+1)={(2n+2)(n+1)\over2}-1-3$
Assuming $a_1=4$, we have $a_n={(2n+2)(n+1)\over2}-1-3+4=(n+1)^2$
A: Calculate $$\sum_{n=0}^{j-1}(a_{n+1} - a_{n}) = \sum_{n=0}^{j-1}(c_{1}n+c_{2})$$
The left side telescopes, to get $$a_j-a_0=\sum_{n=0}^{j-1}(c_{1}n+c_{2})$$
Hence $$a_j=a_0+c_1\sum_{n=0}^{j-1}n+c_2j=a_0+c_1\frac{(j-1)j}{2}+c_2j$$
A: Another way is to make an educated guess.
Notice:
$$F(n+1)-F(n)=g(n) \approx F'(n)$$
Here $g(n)$ is linear so it's integral is quadratic. So we guess a quadratic equation to represent $F$.
If $F(n)=a_n=an^2+bn+c$ then,
$$F(n+1)-F(n)=2an+a+b$$
Now equate your coefficients:
$$2a=2$$
$$a+b=3$$
But really the summation on both sides method is a better way to go.
A: $$\begin{cases}
a_{n+1} - a_{n} = c_1n+c_2 \\
a_{n+2} - a_{n+1} = c_1(n + 1)+c_2 \\
\end{cases}$$
$$a_{n+2} -2a_{n+1} + a_n = c_1$$
Can you take it from here?
