Solving recurrence relation $a_n=a_{n-1} + 3(n-1), a_0=1$ a) $a_n = a_{n-1} + 3(n - 1)$
My work:
Homog. Soln:
$a_n - a_{n-1} = 0$
$\lambda - 1 = 0\Rightarrow \lambda=1$
$a_n = c_1 1^n$
$a_0 = c_1 = 1$
So the homogeneous solution is $a_n = 1$. 
Particular soln:
Guess solution is $Bn + C$. 
$(Bn + C) - (B(n-1) + C) = 3n - 3$
$(Bn - Bn) + (C + B - C) = 3n - 3$
So: $(Bn - Bn) = 3n$ and $(C + B - C) = -3$
But that seems nonsensical, and not sure where to go from there. 
 A: $$a_n-a_{n-1}=3(n-1)$$
$$\sum_{k=1}^n(a_k-a_{k-1})=\sum_{k=1}^n3(k-1)=3(\sum_1^nk-\sum_1^n1)=3n(n+1)/2-3n$$
$$a_n-a_0=3n(n+1)/2-3n$$
$$a_n=1+3n(n+1)/2-3n$$
A: Note that $F(n)=3(n-1)=(3n-3)=(3n-3)(1)^n$. We need to keep in mind the following theorem.

Theorem: Suppose that $\{a_n\}$ satisfies the linear non-homogenous recurrence relation $$a_n=c_1a_{n-1}+c_2a_{n-2}+… c_ka_{n-k} +F(n)$$ where $c_1,c_2,…c_k \in \mathbb{R}$ and $F(n)= (b_tn^t+b_{t-1}n^{t-1}+…b_1n+b_0)s^n$, where $b_0, b_1, …b_t, s \in \mathbb{R}$.
When $s$ is not a root of the characteristic equation of the associated linear homogenous recurrence relation, there is a particular solution of the form $$(p_tn^t + p_{t-1}n^{t-1}+…p_1n+p_0)s^n$$
When $s$ is a root with multiplicity $m$, then the solution is of the form $$n^m(p_tn^t + p_{t-1}n^{t-1}+…p_1n+p_0)s^n$$

Taken from Rosen, Discrete Mathematics & Its Applications.
Coming back to the question, there is thus a particular solution of the form $n(p_1n+p_0)=p_1n^2+p_0n$ as $s=1$ is a root of degree $1$ of the characteristic equation.
Hope you can take it from here.
A: Otherwise you can do the descent $$\begin{array}{ccc}a_n &= &a_{n-1}+3(n-1)\\&=&a_{n-2}+3(n-2)+3(n-1)\\&=&\cdots\\&=&a_1+3\cdot[1+2+3+\cdots+(n-1)]\end{array}$$
A: Another way :
Let $a_n=b_n+p_0+p_1n+p_2n^2+p_3n^3+\cdots$
$$3n-3=a_n-a_{n_1}=b_n-b_{n-1}+p_1+p_2(2n-1)+p_3\{n^3-(n-1)^3\}+\cdots$$
Set $p_r=0\forall r\ge3$
and $2p_1=3,p_1-p_2=-3$ to find $b_n=b_{n-1}$
$$=\cdots=b_0=a_0=?$$
A: Let's keep it simple. The first thing to try is $a_n=f_n+pn+q$. That doesn't work, so take it to $a_n=f_n+pn^2+qn$. You'll readily find that $p=-q=3/2$ and finally, for $a_0=1$ we get
$$a_n=1+\frac{3}{2}n(n-1)$$
A: Your recurrence is particularly simple and the easiest approach seems to be the one suggested by Gerry Myerson. But the standard approach of generating functions works with almost equal ease.
Let $f(x) =\sum_{n=0}^{\infty}a_{n}x^{n}$ so that $$f(x) - xf(x) = a_{0}+\sum_{n=1}^{\infty}(a_{n}-a_{n-1})x^{n}=1+3x^{2}\sum_{n=1}^{\infty}nx^{n-1}=1+\frac{3x^{2}}{(1-x)^{2}}$$ so that $$f(x) =\frac{1}{1-x}+\frac{3x^2}{(1-x)^{3}}$$ Now via binomial theorem we can easily see that $$a_{n} =1+3\cdot\frac{n(n-1)}{2}$$
