I have this
$$a_{n+1} = a_n + 4n – 1\qquad a_1 = 2$$
And I need to find general formula for $a_n$.
This is one of the last exercises for the question related to it so I'll give a summary of what I did before because maybe it could be needed for this.
$$b_n = 2n^2 + 2n – a_n$$
I found that $b_n$ is an arithmetic progression and that $d_b = 5$ and that $$b_n = 5n-3$$
What I have tried for finding the formula is:
If it is an arithmetic progression then:
$$d_a = a_{n+1} – a_n$$
But I get that $d_a = 4n-1$ so it's not good, it's not the same $d$ for always.
Then I tried as geometric progression:
$$r_a = \frac{a_{n+1}}{a_n} = \frac{a_n + 4n – 1}{a_n} $$
So here I got stuck. (By the way, these are the type of series I have learned)
Best Answer
For the non-homogeneous part try $b_n=cn^2+dn+e$ so $$ b_{n+1}=b_n+4n-1\quad\Longrightarrow\quad c(n+1)^2+d(n+1)+e=cn^2+dn+ e+4n-1\\ cn^2+2cn+c+dn+d+e=cn^2+dn+ e+4n-1\\ 2cn+c+d=4n-1 $$ and then $c=2$ and $d=-3$.
For the homogeneous part $a_{n+1}=a_n$ so that $a_n=k$.
So the general solution is $$ a_n=k+2n^2-3n $$ From the initial condition $a_1=2$ we have $2=k+2-3$ and then $k=3$
So the final solution is $$ a_n=3+2n^2-3n $$