Can you find the general formula for A_n+3?

[kezman]

Hi I have many problems trying to find the general formula and the demonstration by induction.
Let
$$A_1 = 3$$
$$A_2 = 7$$
$$A_3 = 13$$
A_n+3 = 3A_n+2 - 3A_n+1 + A_n
I could only find this.
A_n+1 = A_n + 2n

 

[Hurkyl]

What sort of class are you in?

A_n+1 = A_n + 2n

How did you arrive at this? It can't be right, though: it doesn't work for n=1 or n=2.

Oh, I bet you meant $A_n = A_{n-1} + 2n$, or $A_{n+1} = A_n + 2(n+1)$. (Incidentally, you could try proving this formula by induction, to make sure you're on the right track)


One trick that's often useful is to not do arithmetic. If $A_n = A_{n-1} + 2n$, then write $A_2 = 3 + 2\cdot2$ instead of $A_2 = 7$.

 

[kezman]

sorry I am in my way of learning Latex.
I mean:
$A_n = A_{n-1} + 2n$
And the formula given in the problem is:
$A_{n+3} = 3A_{n+2} - 3A_{n+1} + A_n$
With the three A1 A2 A3 defined.
Ive been trying but I couldn't find the general formula.

 

[VietDao29]

...I mean:
$A_n = A_{n-1} + 2n$

So far so good.
Now, you can follow Hurkyl's suggestion:
A₁ = 3
A₂ = A₁ + 2 . 2 = 3 + 2 . 2
A₃ = A₂ + 2 . 3 = 3 + 2 . 2 + 2 . 3
A₄ = 3 + 2 . 2 + 2 . 3 + 2 . 4
A₅ = 3 + 2 . 2 + 2 . 3 + 2 . 4 + 2 . 5
A₆ = 3 + 2 . 2 + 2 . 3 + 2 . 4 + 2 . 5 + 2 . 6
...
A_n = 3 + 2 . 2 + 2 . 3 + 2 . 4 + 2 . 5 + 2 . 6 + 2 . 7 + ... + 2 . n
Now, do you see anything that can be factored out?
Can you go from here? :)

 

[kezman]

For
$A_n = 3 + \sum\limits_{i = 2}^n 2i$
then
$A_1 = 3$
Is this correct?

 

[Hurkyl]

If you can prove it by induction, it's correct.

 

[kezman]

Induction:
1)
$A_1 = 3 + \sum\limits_{i = 2}^1 2i = 3$
(Is this OK?)
2) I had this result :$A_{n+1} = A_n + 2(n+1)$

With $A_n = 3 + \sum\limits_{i = 2}^n 2i$ =>
$A_{n+1} = 3 + \sum\limits_{i = 2}^{n+1} 2i$ (the upper index of the summatory in this case is (n+1))
=> $A_{n+1} = 3 + \sum\limits_{i = 2}^n 2i + 2(n+1)$
=> $A_{n+1} = A_n + 2(n+1)$

Is this correct?

 

[Hurkyl]

Well, you've done it backwards!

$$A_{n+1} = A_n + 2(n+1)$$

is the statement you know to be true, whereas

$$A_{n+1} = 3 + \sum\limits_{i = 2}^{n+1} 2i$$

is the statement you were trying to prove.

(given the assumption that $A_n = 3 + \sum_{i = 2}^n 2i$)



(Actually, I'm assuming you've already given a proof of $A_{n+1} = A_n + 2(n+1)$ is correct given the original recurrence relation. If you have not yet done so, then you should work with the original recurrence)

 

[kezman]

Yes is true.I still have to prove first $$A_{n+1} = A_n + 2(n+1)$$