Solving a series with this form without using induction or countdown method

Question

Ok so i have to solve the question above and find the general term of this sequence without using the induction method or the countdown method. So i want to calcul this with the geometric formula or arithmethic given :

I thought this would be a summation of an arithmetic sequence but using the formula i don't get the answer.

I've tested the answer with wolframalpha entering : y(0)=2,y(n)=2y(n-1)+3n-4

Using the geometric formula i get close to the answer but missing -3n, so i'm kinda lost. Could anyone help thanks !

Maybe it can't be solved with geometric or arithmetic formula and i'd have to use an other method ?

Answer 1

Let $b_m=a_m+p+qm$

$3n-4=a_n-2a_{n-1}+p+qn-2(p+(q-1)n)$

Set $p-2p=-4$ and $2-q=3$ to find $a_n=2a_{n-1}=2^ra_{n-r}$

Answer 2

Assume $a_n=b_n+pn+q$ therefore $$$$

therefore $$a_n=b_n+3n+2$$therefore $$a_{n}{=b_{n}+pn+q\\=2b_{n-1}+3n-4+pn+q\\=(3+p)n+q-4+2(a_{n-1}-pn+p-q)\\=2a_{n-1}+n(3-p)+2p-4-q}$$by defining $p=3$ and $q=2$ we obtain $$a_n=2a_{n-1}$$ and since $a_0=b_0+2=4$ we conclude that $a_n=2^{n+2}$ and$$b_n=2^{n+2}-3n-2$$