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.

2. 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:

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

number ending in 1 ends in 9...

Tonio

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

4. Originally Posted by Plato
If you want to do it by induction here is a trick.
$3^{4(n+1)+2}+1=3^{4(n+1)+2}+3^4-3^4+1=3^4(3^{4n+2}+1)-3^4+1$
So how does that show its divisible by 10? im a little confused

5. 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.

6. Wonder why the topic says "Proof by contradiction " ?? By induction you mean ?