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)