Proof by contradiction

**kensington** · February 27th 2011, 02:40 PM

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.

**tonio** · February 27th 2011, 02:56 PM

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.

Click image for larger version.

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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

**Plato** · February 27th 2011, 03:23 PM

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

**kensington** · February 27th 2011, 03:54 PM

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

**DrSteve** · February 27th 2011, 04:33 PM

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.

**Tarask** · February 28th 2011, 10:06 AM

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

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