Finding general formula for sequence

Question

I posted a question like this one but didn't understand no one of the answers! Sometimes it was a bit too complicated, they used $\Sigma$ that I didn't learn, and some I didn't understand what they meant! If you can give very simple answers or explain slowly each step I will appreciate it.

I have $$a_{n+1} = a_n + 4n - 1\qquad a_1 = 2$$

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)

Answer 1

Your recurrence reads $$a_{n+1}-a_n=4n-1$$ That is, the sequence of differences is a linear polynomial. That is enormously informative! Quadratics have this property. Now, in fact it's true that only quadratics have this property but we don't need to prove that theorem. It certainly sounds plausible, at least plausible enough to guess that our solution must be of the form $a_n=An^2+Bn+C$. Let's try to solve for $A,B,C$. Using our guessed form we get: $$a_{n+1}-a_n=A(n+1)^2+B(n+1)+C-An^2-Bn-C=2An+A+B=4n-1$$ This immediately tells us that $$2A=4\;\;\&\;\;A+B=-1\implies A=2\;\;\&\;\;B=-3$$ Now you also specify the initial condition $$a_1=2\implies A+B+C=2\implies -1+C=2\implies C=3$$ Thus our answer is $$a_n=2n^2-3n+3$$

Answer 2

WLOG let $a_m=p_m+am^2+bm+c$

$4n-2=a_n-a_{n-1}=p_n-p_{n-1}+a(2n-1)+b$

Set $a=2, b=1$ to find $p_n=p_{n-1}=\cdots=p_1$

But $m=1\implies2=a_1=p_1+a+b+c\iff p_1=2-a-b-c=-1-c$

Answer 3

$$a_m=a_{m-1}+4m-1$$

Put $m=2,3,\cdots, n-1,n$ and add to get

$$a_n-a_1=\sum_{m=2}^n(4m-1)=\dfrac{n-1}2\left(4\cdot2-1+4n-1\right)$$