In article
af173af3-bff0-489b-b090-4a8854148...@r66g2000hsg.googlegroups.com>,
Gerard Matthew <calibis...@gmail.com> wrote:
What do you mean?
On May 1, 11:58?pm, William Hale <h...@tulane.edu> wrote:
In article
0be2cde6-535f-4f34-99eb-825aa1eeb...@x35g2000hsb.googlegroups.com>,
?Gerard Matthew <calibis...@gmail.com> wrote:
Prove that 5 | (3^(3n+1) + 2^(n+1)) for every positive n.
What I did was a proof by induction.
Basically
Let n =0, 5 | 3 + 2 is ture
Assume ?5 | (3^(3(k+1)+1) + 2^((k+1)+1))
(3^(3(k+1)+1) + 2^((k+1)+1)) = (3^3k).(3^3) + (2^k).(2^2)
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 27(3^3k) +
4(2^k)
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 81.3^k + 4.2^k
I know that I have to prove that the above is divisible by 5 but i'm
lost as to how to go about doing that?
You didn't make use of the induction assumption:
? ? ?5 divides 3^(3k+1) + 2^(k+1)
To prove that 5 divides 81.3^k + 4.2^k, you can use the induction
hypothesis that 5 divides 3^(3k+1) + 2^(k+1).
Try to work 81.3^k + 4.2^k into a form like 3^(3k+1) + 2^(k+1).
For example, that 4.2^k doesn't have (k+1) as exponent of the 2.
How would you rewrite 4.2^k so that it has (k+1) as exponent of 2?
