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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)

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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 $$ a_n=3+2n^2-3n $$

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$$\begin{align} a_{n+1}&=a_n+4n-1\\ a_{n+1}-a_n&=4n-1\\ a_n-a_{n-1}&=4(n-1)-1\\ &\vdots\\ a_2-\underbrace{a_1}_{=2}&=4(1)-1\\ \end{align}$$ Summing the last $(n-1)$ lines by telescoping gives $$\begin{align} a_{n}-2&=4\cdot \frac {n(n-1)}2-(n-1)\\ a_n&=2n^2-3n+3\qquad\blacksquare\end{align}$$

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As you tried, $(a_n)$ is neither arithmetic nor geometric. However, \begin{align} a_n &= \sum_{i=1}^{n-1} (a_{i+1}-a_i) + a_1\\ &=\sum_{i=1}^{n-1}(4i-1) +2\\ &=2n(n-1)-(n-1)+2\\ &=2n^2-3n+3. \end{align}

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  • $\begingroup$ Could you edit this without using $\Sigma$? I just didn't learn about it (I just know as general knowledge what it is and what it does) so it's harder to understand it. $\endgroup$ – Pichi Wuana Apr 8 '16 at 21:48
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if you knew about sequences of differences, you can also use that.

The first differences will be 4n-1 . The second differences are all 4. Using "You can simplify your computations somewhat by using a formula for the leading coefficient of the sequence's polynomial. The coefficient of the first term of the polynomial will be equal to the common difference divided by the factorial of the polynomial's degree." from purplemath.com, we get the leading coefficient is $\frac{4}{2!}=2$, so subtract $2(n+1)^2$ from each index. The difference, is then -3(n+1). So, we would do that subtraction again and get a common value of 3. So $2n^2-3n+3$ is the value at index n.

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