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by **hbelle** on Thu Mar 10, 2011 4:17 pm

What is the closed form for recursive equation [a][/0] = 0, [a][/1] = 1 and [a][/n] = [3a][/n-1] - [2a][/n-2]

is the closed form [a][/k+1] = [3a][/k+1-1] - [2a][/k+1-2] ???

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by **stapel_eliz** on Thu Mar 10, 2011 6:19 pm

hbelle wrote:What is the closed form for recursive equation [a][/0] = 0, [a][/1] = 1 and [a][/n] = [3a][/n-1] - [2a][/n-2]

What is the definition of the square brackets and the "slash" notation?

Thank you! :wink:

by **Martingale** on Sat Mar 12, 2011 5:21 am

hbelle wrote:What is the closed form for recursive equation [a][/0] = 0, [a][/1] = 1 and [a][/n] = [3a][/n-1] - [2a][/n-2]

is the closed form [a][/k+1] = [3a][/k+1-1] - [2a][/k+1-2] ???

if you had

$a_0 = 0, a_1 = 1 \text{ and } a_n = 3a_{n-1} - 2a_{n-2}$

then the closed form is just

$a_n=2^n-1$

The Purplemath Forums

Closed Form of ...[a][/n] = [3a][/n-1] - [2a][/n-2]

Closed Form of ...[a][/n] = [3a][/n-1] - [2a][/n-2]

Re: Closed Form of ...[a][/n] = [3a][/n-1] - [2a][/n-2]