Recursive Relation: a_0 = 2, a_1 = 5, a_n = 21a_(n-2), ....

D

deemanw

New member
Joined
Aug 2, 2006
Messages
11
find nth term formula and prove result by induction for: a sub(n)=21a sub(n-2) - 4a sub(n-1):n>=2; a sub(0) = 2; a sub(1) = 5.

so far i got 2+5+(7*2*3 - 2*2*5) + (3*7*5 - 4*2*11) + (3*7*2*11 - 2*2*17) ....... and i'm trying to figure how to make the formula???? any help would be greatly appreciated, thanks.

my induction skils are not so good either, thanks
 
Is it
\(\displaystyle \
a_n = 21a_{n - 2} - 4a_{n - 1} \quad \mbox{or} \quad a_n = 2a_{n - 2} - 4a_{n - 1} \quad \mbox{or what?} \quad\)

I cannot follow your work.
You ought learn to use LaTeX.
 
hey thanks for the quick response, and yes i will learn latex when i can. the answer to your question is the former. thanks for your help! :(
 
Well I know how to solve the linear homogeneous recursion.
I can tell you the correct answer, but not how to do it.
\(\displaystyle a_n = \left( {0.1} \right)\left( { - 7} \right)^n + \left( {1.9} \right)\left( 3 \right)^n.\)
You can find the technique in any standard Discrete Mathematics text.
 
deemanw said:
what does the vertical bar in the middle mean? "divides"?
Click to expand...
What vertical bar?? There is no vertical bar!
Do you mean \(\displaystyle \(- 7\)\)? That is minus seven.
 
You must log in or register to reply here.
Top