induction on a reccurence sequence

Printable View

November 29th 2008, 07:28 AM
ad53ggz

induction on a reccurence sequence

i have a1=1

a(n+1) = (5an+4)/(an+2) for n>=1

need to prove by induction on n that 0<an<4 .

course of action would be?

then n is subscript and (n+1) also subscript , i think i need to get it into the form an+2 =A an+1 + B an

where A and B are some fixed numbers?

thanks
November 29th 2008, 08:03 AM
running-gag

Hello !

The start of the recurrence is OK since 0 < a1=1 < 4

Suppose that $0 < a_n < 4$
Then
$a_{n+1}\;=\;\frac{5a_n+4}{a_n+2}$

$a_{n+1}\;=\;5-\frac{6}{a_n+2}$

$0 < a_n < 4$

$2 < a_n +2 < 6$

$\frac{1}{6} < \frac{1}{a_n+2} < \frac{1}{2}$

$1 < \frac{6}{a_n+2} < 3$

$2 < 5-\frac{6}{a_n+2} < 4$

$0 < 2 < a_{n+1} < 4$
November 29th 2008, 08:24 AM
ad53ggz

an+1 = 5-6/an+2

spot on mate was this bit i couldnt work out feel dumb now (Giggle)

All times are GMT -8. The time now is 01:51 AM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.