Prove by Induction: 3n-1 is Divisible by 2

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The discussion focuses on proving by induction that the expression 3n-1 is divisible by 2. The initial base case is established with n=1, where 3-1 equals 2, confirming divisibility. The induction hypothesis is defined as 3^n = 2a + 1 for some integer a. By substituting this hypothesis into the expression for 3^{n+1}-1, the proof concludes that 6a + 2 can be factored as 2(3a + 1), demonstrating that it is indeed divisible by 2.

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BloodyFrozen
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I was doing this for practice and came across this problem-- I have no clue how to prove it.

By Induction, prove that 3n-1 is divisible by 2.

1) 3-1=2, divisible by two; good so far

Now I have no clue how to approach this.:bugeye:

Please help. Thanks in advance:smile:
 
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Hi BloodyFrozen! :smile:

So, the induction hypothesis says that

3^n=2a+1

for some a. Now, try substitute this in 3^{n+1}-1.
 
micromass said:
Hi BloodyFrozen! :smile:

So, the induction hypothesis says that

3^n=2a+1

for some a. Now, try substitute this in 3^{n+1}-1.


What do you mean sustituting it in?
 
BloodyFrozen said:
What do you mean sustituting it in?

3^{n+1}-1=3.3^n-1=3.(2a+1)-1
 
micromass said:
3^{n+1}-1=3.3^n-1=3.(2a+1)-1

Ok.
6a+2=2(3a+1).

Since this is multiplied by 2, it must be divisible by 2.
Thanks again (if you know what I mean ;P) micromass!
 

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