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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)

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  • $\begingroup$ Do you know how to use Undetermined Coefficients (a.k.a. "inspired guesswork")? $\endgroup$ – lulu Apr 9 '16 at 10:48
  • $\begingroup$ @lulu The only thing that comes to my mind is something related to trial and error...... $\endgroup$ – Pichi Wuana Apr 9 '16 at 10:49
  • $\begingroup$ I'll write up a quick argument and post it below. Tell me if it helps. $\endgroup$ – lulu Apr 9 '16 at 10:50
  • $\begingroup$ Okay great thanks! $\endgroup$ – Pichi Wuana Apr 9 '16 at 10:53
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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 $$a_n=2n^2-3n+3$$

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  • $\begingroup$ Wow it worked! I would never think about this. Thanks! $\endgroup$ – Pichi Wuana Apr 9 '16 at 11:06
  • $\begingroup$ Glad it helped. It's a very useful, and general, technique...if you can guess what the form of the answer might be you can then try to nail it down. In the process you'll either solve the problem (as here) or you'll prove that the form you thought of was wrong (which at least rules out some possibilities). $\endgroup$ – lulu Apr 9 '16 at 11:11
  • $\begingroup$ What can be the next try? $\endgroup$ – Pichi Wuana Apr 9 '16 at 11:12
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    $\begingroup$ Depends on the situation. For polynomials, the answer is always just a polynomial one degree higher. In general, you try to think of similar problems with known solutions and you guess that the answer will also be similar. To be clear: this does not always work. But even when it fails you usually gain some insight. $\endgroup$ – lulu Apr 9 '16 at 11:14
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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 $m=1\implies2=a_1=p_1+a+b+c\iff p_1=2-a-b-c=-1-c$

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

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

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

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i believe this way also it can be solved

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