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
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].
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.
(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)
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