• Feb 27th 2011, 02:40 PM
kensington
I am having difficulty with the following question: ( see attachment)
I tried expanding it and i still didnt seem to quite get it.
Any help would be apriciate, thanks in advance. :)
• Feb 27th 2011, 02:56 PM
tonio
Quote:

Originally Posted by kensington
I am having difficulty with the following question: ( see attachment)
I tried expanding it and i still didnt seem to quite get it.
Any help would be apriciate, thanks in advance. :)

Induction on n is a straighforward way, but you can do as follows as well:

$\displaystyle 3^{4n+2}+1=9\cdot (3^4)^n+1=9\cdot (81)^n+1$ , and you're done since 9 times a

number ending in 1 ends in 9...

Tonio
• Feb 27th 2011, 03:23 PM
Plato
If you want to do it by induction here is a trick.
$\displaystyle 3^{4(n+1)+2}+1=3^{4(n+1)+2}+3^4-3^4+1=3^4(3^{4n+2}+1)-3^4+1$
• Feb 27th 2011, 03:54 PM
kensington
Quote:

Originally Posted by Plato
If you want to do it by induction here is a trick.
$\displaystyle 3^{4(n+1)+2}+1=3^{4(n+1)+2}+3^4-3^4+1=3^4(3^{4n+2}+1)-3^4+1$

So how does that show its divisible by 10? im a little confused
• Feb 27th 2011, 04:33 PM
DrSteve
Quote:

Originally Posted by kensington
So how does that show its divisible by 10? im a little confused

The first term is divisible by 10 by the inducutive hypothesis, and you can just add up the last 2 terms to see that it ends in a 0.
• Feb 28th 2011, 10:06 AM