Find the general formula (1 Viewer)

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

Hurkyl

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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 [itex]A_n = A_{n-1} + 2n[/itex], or [itex]A_{n+1} = A_n + 2(n+1)[/itex]. (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 [itex]A_n = A_{n-1} + 2n[/itex], then write [itex]A_2 = 3 + 2\cdot2[/itex] instead of [itex]A_2 = 7[/itex].
 
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sorry im in my way of learning Latex.
I mean:
[itex]A_n = A_{n-1} + 2n[/itex]
And the formula given in the problem is:
[itex]A_{n+3} = 3A_{n+2} - 3A_{n+1} + A_n[/itex]
With the three A1 A2 A3 defined.
Ive been trying but I couldnt find the general formula.
 
Last edited:

VietDao29

Homework Helper
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kezman said:
...I mean:
[itex]A_n = A_{n-1} + 2n[/itex]
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? :)
 
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For
[itex] A_n = 3 + \sum\limits_{i = 2}^n 2i [/itex]
then
[itex] A_1 = 3 [/itex]
Is this correct?
 
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Hurkyl

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If you can prove it by induction, it's correct. :smile:
 
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Induction:
1)
[itex] A_1 = 3 + \sum\limits_{i = 2}^1 2i = 3 [/itex]
(Is this OK?)
2) I had this result :[itex]A_{n+1} = A_n + 2(n+1)[/itex]

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

Is this correct?
 
Last edited:

Hurkyl

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Well, you've done it backwards!

[tex] A_{n+1} = A_n + 2(n+1) [/tex]

is the statement you know to be true, whereas

[tex]A_{n+1} = 3 + \sum\limits_{i = 2}^{n+1} 2i[/tex]

is the statement you were trying to prove.

(given the assumption that [itex] A_n = 3 + \sum_{i = 2}^n 2i [/itex])



(Actually, I'm assuming you've already given a proof of [itex] A_{n+1} = A_n + 2(n+1) [/itex] is correct given the original recurrence relation. If you have not yet done so, then you should work with the original recurrence)
 
37
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Yes is true.I still have to prove first [tex] A_{n+1} = A_n + 2(n+1) [/tex]
 

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