# Find the general formula

1. Apr 22, 2006

### 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

Last edited: Apr 22, 2006
2. Apr 22, 2006

### Hurkyl

Staff Emeritus
What sort of class are you in?

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 $A_2 = 7$.

3. Apr 22, 2006

### kezman

sorry im 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 couldnt find the general formula.

Last edited: Apr 22, 2006
4. Apr 22, 2006

### VietDao29

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
...
An = 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? :)

5. Apr 22, 2006

### kezman

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

Last edited: Apr 23, 2006
6. Apr 22, 2006

### Hurkyl

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

7. Apr 23, 2006

### 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?

Last edited: Apr 23, 2006
8. Apr 23, 2006

### Hurkyl

Staff Emeritus
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)

9. Apr 23, 2006

### kezman

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