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PhoenixFire
 3 years ago
Using mathematical induction prove that for \[ {n \ge 1} \in \mathbb{Z} \]
\[ 3^{2k}1 \] is divisible by 8.
PhoenixFire
 3 years ago
Using mathematical induction prove that for \[ {n \ge 1} \in \mathbb{Z} \] \[ 3^{2k}1 \] is divisible by 8.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0show it for certain number n ... show it hods for n+1

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah I got the base step. n=1 \[3^21=8k\] where k is some Integer. \[8=8k\] so it's divisible.

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Then the inductive hypothesis is: Assume \[3^{2k}1\] is divisible by 8 when k=n Now to prove it hold for k=n+1, I get confused and don't know what to do.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is difficult. let's try without induction first. you have to show that there is 3 even factors.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.03^(2k)  1 = (3^k)^2  1^2 = (3^k  1)(3^k + 1) < one of them is divisible by 4. and both is divisible by 2 ... hence it is divisible by 8

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, I understand that. Now what?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i don't see short sweet method.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Prove ... if (3^k  1) is divisible by 2 prove (3^k + 1) is always divisible by 4 using induction. and viceversa ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you want to show that give \(3^{2k}1\) is divisible by \(8\) then \(3^{2(k+1)}1\) is as well

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that is, if it is true for \(n=k\) then it is true for \(k+1\) now some algebra slight of hand

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.03^(2(k+1))  1 = 9(3^(2k)  1) + 1)  1 = 9(x+1)  1 = 9*8x + 8 which is divisible by 8

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and the goal is to get \(3^{2k}1\) somehow out of the exression \(3^{2(k+1)}1\) so that you can use the "induction hypothesis" of the term \(3^{2k}1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ley 8x = 3^(2k)  1 3^(2(k+1))  1 = 9 ( (3^(2k)  1) + 1)  1 = 9( 8x + 1 )  1 = 9*8x + 8 < hence it is divisible by 8

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.0\[k = n\] \[3^{2n}  1 = 8Q (where ~Q~ is~ any~ integer) => 3^{2n} = 8Q+1\] \[k= n+1 \] \[3^{2(n+1)}  1 => 3^{2n+1}  1 => 3^{2n} *3  1 => \left(8Q+1\right) *3 1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[3^{2(k+1)}1=3^{2k+2}1=3^{2k}3^21=9\times 3^{2k}1\] \[=(3^{2k}1)\times 9+8\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by induction, the terms \(3^{2k}1\) is divisible by \(8\) and clearly \(9\times 8\) is divisible by \(8\) so the sum is divisible by \(8\)

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, I now understand, however I have one question... Where the heck did the +8 come from? You have \[(9)3^{2k}1\] how did that become \[9(3^{2k}1)+8\] Actually, I understand @RaphaelFilgueiras Is it much different? @satellite73 or am I just missing something in the working.

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks everyone! I've got it now.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.03^(2k) = (3^(2k)  1) + 1 < 8 comes from here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is a trick of algebra, and the reason to use the trick is to make sure you can get \(3^{2k}1\) out of \(3^{2k+2}1\) so you can use the induction hypothesis that is how i arrived at it to being with i just wrote \(3^{2k}\times 91\) and forced \(3^{2k}1\) out of it
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